31:T8e4,md`

--- Day 4: Passport Processing ---

You arrive at the airport only to realize that you grabbed your North Pole Credentials instead of your passport. While these documents are extremely similar, North Pole Credentials aren't issued by a country and therefore aren't actually valid documentation for travel in most of the world. It seems like you're not the only one having problems, though; a very long line has formed for the automatic passport scanners, and the delay could upset your travel itinerary. Due to some questionable network security, you realize you might be able to solve both of these problems at the same time. The automatic passport scanners are slow because they're having trouble detecting which passports have all required fields. The expected fields are as follows: byr (Birth Year) iyr (Issue Year) eyr (Expiration Year) hgt (Height) hcl (Hair Color) ecl (Eye Color) pid (Passport ID) cid (Country ID) Passport data is validated in batch files (your puzzle input). Each passport is represented as a sequence of key:value pairs separated by spaces or newlines. Passports are separated by blank lines. Here is an example batch file containing four passports: ecl:gry pid:860033327 eyr:2020 hcl:#fffffd byr:1937 iyr:2017 cid:147 hgt:183cm iyr:2013 ecl:amb cid:350 eyr:2023 pid:028048884 hcl:#cfa07d byr:1929 hcl:#ae17e1 iyr:2013 eyr:2024 ecl:brn pid:760753108 byr:1931 hgt:179cm hcl:#cfa07d eyr:2025 pid:166559648 iyr:2011 ecl:brn hgt:59in The first passport is valid - all eight fields are present. The second passport is invalid - it is missing hgt (the Height field). The third passport is interesting; the only missing field is cid, so it looks like data from North Pole Credentials, not a passport at all! Surely, nobody would mind if you made the system temporarily ignore missing cid fields. Treat this "passport" as valid. The fourth passport is missing two fields, cid and byr. Missing cid is fine, but missing any other field is not, so this passport is invalid. According to the above rules, your improved system would report 2 valid passports. Count the number of valid passports - those that have all required fields. Treat cid as optional. In your batch file, how many passports are valid?

`32:T4b9,// part 2 - implement field validation mdPlus`--- Part Two --- The line is moving more quickly now, but you overhear airport security talking about how passports with invalid data are getting through. Better add some data validation, quick! You can continue to ignore the cid field, but each other field has strict rules about what values are valid for automatic validation: - [x] byr (Birth Year) - four digits; at least 1920 and at most 2002. - [x] iyr (Issue Year) - four digits; at least 2010 and at most 2020. - [x] eyr (Expiration Year) - four digits; at least 2020 and at most 2030. - [x] hgt (Height) - a number followed by either cm or in: * If cm, the number must be at least 150 and at most 193. * If in, the number must be at least 59 and at most 76. - [x] hcl (Hair Color) - a # followed by exactly six characters 0-9 or a-f. - [x] ecl (Eye Color) - exactly one of: amb blu brn gry grn hzl oth. - [x] pid (Passport ID) - a nine-digit number, including leading zeroes. - [x] cid (Country ID) - ignored, missing or not. Your job is to count the passports where all required fields are both present and valid according to the above rules. Here are some example values:`33:T4c4,function day5BoardingPass(id) { const row = (boardingPass => { // array of operations, limit to F or B chars let operations = boardingPass .split("") .filter(op => op === "F" || op === "B"); let min = 0, max = 127; let operation = operations.shift(); while (operation && min !== max) { const delta = (max - min + 1) / 2; switch (operation) { case "F": max -= delta; break; case "B": min += delta; break; } operation = operations.shift(); } return max; })(id); const col = (boardingPass => { // array of operations, limit to F or B chars let operations = boardingPass .split("") .filter(op => op === "L" || op === "R"); let min = 0, max = 7; let operation = operations.shift(); while (operation && min !== max) { const delta = (max - min + 1) / 2; switch (operation) { case "L": max -= delta; break; case "R": min += delta; break; } operation = operations.shift(); } return max; })(id); return { row, col, id: row * 8 + col }; }34:Ta8c,md`

--- Day 5: Binary Boarding ---

You board your plane only to discover a new problem: you dropped your boarding pass! You aren't sure which seat is yours, and all of the flight attendants are busy with the flood of people that suddenly made it through passport control.

You write a quick program to use your phone's camera to scan all of the nearby boarding passes (your puzzle input); perhaps you can find your seat through process of elimination.

Instead of zones or groups, this airline uses binary space partitioning to seat people. A seat might be specified like FBFBBFFRLR, where F means "front", B means "back", L means "left", and R means "right".

The first 7 characters will either be F or B; these specify exactly one of the 128 rows on the plane (numbered 0 through 127). Each letter tells you which half of a region the given seat is in. Start with the whole list of rows; the first letter indicates whether the seat is in the front (0 through 63) or the back (64 through 127). The next letter indicates which half of that region the seat is in, and so on until you're left with exactly one row.

For example, consider just the first seven characters of FBFBBFFRLR:

Start by considering the whole range, rows 0 through 127.
F means to take the lower half, keeping rows 0 through 63.
B means to take the upper half, keeping rows 32 through 63.
F means to take the lower half, keeping rows 32 through 47.
B means to take the upper half, keeping rows 40 through 47.
B keeps rows 44 through 47.
F keeps rows 44 through 45.
The final F keeps the lower of the two, row 44.
The last three characters will be either L or R; these specify exactly one of the 8 columns of seats on the plane (numbered 0 through 7). The same process as above proceeds again, this time with only three steps. L means to keep the lower half, while R means to keep the upper half.

For example, consider just the last 3 characters of FBFBBFFRLR:

Start by considering the whole range, columns 0 through 7.
R means to take the upper half, keeping columns 4 through 7.
L means to take the lower half, keeping columns 4 through 5.
The final R keeps the upper of the two, column 5.
So, decoding FBFBBFFRLR reveals that it is the seat at row 44, column 5.

Every seat also has a unique seat ID: multiply the row by 8, then add the column. In this example, the seat has ID 44 * 8 + 5 = 357.

Here are some other boarding passes:

BFFFBBFRRR: row 70, column 7, seat ID 567.
FFFBBBFRRR: row 14, column 7, seat ID 119.
BBFFBBFRLL: row 102, column 4, seat ID 820.
As a sanity check, look through your list of boarding passes. What is the highest seat ID on a boarding pass?

`35:T75b,md`

--- Day 6: Custom Customs ---

As your flight approaches the regional airport where you'll switch to a much larger plane, customs declaration forms are distributed to the passengers.

The form asks a series of 26 yes-or-no questions marked a through z. All you need to do is identify the questions for which anyone in your group answers "yes". Since your group is just you, this doesn't take very long.

However, the person sitting next to you seems to be experiencing a language barrier and asks if you can help. For each of the people in their group, you write down the questions for which they answer "yes", one per line. For example:

abcx
abcy
abcz

In this group, there are 6 questions to which anyone answered "yes": a, b, c, x, y, and z. (Duplicate answers to the same question don't count extra; each question counts at most once.)

Another group asks for your help, then another, and eventually you've collected answers from every group on the plane (your puzzle input). Each group's answers are separated by a blank line, and within each group, each person's answers are on a single line. For example:

abc

a
b
c

ab
ac

a
a
a
a

b
This list represents answers from five groups:

The first group contains one person who answered "yes" to 3 questions: a, b, and c.
The second group contains three people; combined, they answered "yes" to 3 questions: a, b, and c.
The third group contains two people; combined, they answered "yes" to 3 questions: a, b, and c.
The fourth group contains four people; combined, they answered "yes" to only 1 question, a.
The last group contains one person who answered "yes" to only 1 question, b.
In this example, the sum of these counts is 3 + 3 + 3 + 1 + 1 = 11.

For each group, count the number of questions to which anyone answered "yes". What is the sum of those counts?

`36:T8f2,md`

--- Day 7: Handy Haversacks ---

You land at the regional airport in time for your next flight. In fact, it looks like you'll even have time to grab some food: all flights are currently delayed due to issues in luggage processing.

Due to recent aviation regulations, many rules (your puzzle input) are being enforced about bags and their contents; bags must be color-coded and must contain specific quantities of other color-coded bags. Apparently, nobody responsible for these regulations considered how long they would take to enforce!

For example, consider the following rules:

light red bags contain 1 bright white bag, 2 muted yellow bags.
dark orange bags contain 3 bright white bags, 4 muted yellow bags.
bright white bags contain 1 shiny gold bag.
muted yellow bags contain 2 shiny gold bags, 9 faded blue bags.
shiny gold bags contain 1 dark olive bag, 2 vibrant plum bags.
dark olive bags contain 3 faded blue bags, 4 dotted black bags.
vibrant plum bags contain 5 faded blue bags, 6 dotted black bags.
faded blue bags contain no other bags.
dotted black bags contain no other bags.
These rules specify the required contents for 9 bag types. In this example, every faded blue bag is empty, every vibrant plum bag contains 11 bags (5 faded blue and 6 dotted black), and so on.

You have a shiny gold bag. If you wanted to carry it in at least one other bag, how many different bag colors would be valid for the outermost bag? (In other words: how many colors can, eventually, contain at least one shiny gold bag?)

In the above rules, the following options would be available to you:

A bright white bag, which can hold your shiny gold bag directly.
A muted yellow bag, which can hold your shiny gold bag directly, plus some other bags.
A dark orange bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
A light red bag, which can hold bright white and muted yellow bags, either of which could then hold your shiny gold bag.
So, in this example, the number of bag colors that can eventually contain at least one shiny gold bag is 4.

How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)

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Naturally, I'll be leaning on JS for the programming exercises, which makes them a good fit for observable.\n\nI'll include the text of the exercise, but mostly focus on the deliverable of each part.\n`","pinned":false,"hidden":false,"mode":"ojs"},{"id":86,"value":"toc({ selector: \"h2,h3,h4,h5,h6\", heading: \"Jump To:\" })","pinned":false,"hidden":false,"mode":"ojs"},{"id":31,"value":"md`## Day 1`","pinned":false,"hidden":false,"mode":"ojs"},{"id":42,"value":"day1input = {\n const text = await FileAttachment(\"day 1.txt\").text();\n return (\n text\n // get each line\n .split(\"\\n\")\n // trim trailing newline, which would otherwise convert to \"0\"\n .filter(e => !!e)\n // coerce numeric\n .map(e => +e)\n );\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":39,"value":"// Part 1: given a text list of numbers, find the two entries whose sum is 2020","pinned":true,"hidden":false,"mode":"ojs"},{"id":48,"value":"day1input\n // grab entries\n .filter((e, i, a) => a.includes(2020 - e))\n // multiply\n .reduce((a, e) => a * e)","pinned":true,"hidden":false,"mode":"ojs"},{"id":40,"value":"////////////////////\n// Part 2: given that same list of items, find the three entries whose sum is 2020","pinned":true,"hidden":false,"mode":"ojs"},{"id":57,"value":"{\n const numbers = day1input;\n const collection = [];\n\n // easy way to implement two cursors\n for (let i of numbers) {\n for (let j of numbers) {\n if (numbers.includes(2020 - i - j)) {\n const v = [i, j, 2020 - i - j].sort();\n collection.push({ v, total: v.reduce((a, e) => a * e) });\n }\n }\n }\n return JSON.stringify(collection[0]);\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":36,"value":"md`## Day 2`","pinned":false,"hidden":false,"mode":"ojs"},{"id":69,"value":"day2input = {\n const input = await FileAttachment(\"day 2@1.txt\").text();\n return input\n .split(\"\\n\")\n .filter(line => line !== \"\")\n .map(line => line.split(/[\\ \\-\\:]/));\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":63,"value":"// Part 1: given a list of passwords, how many have the character the number of instances?","pinned":true,"hidden":false,"mode":"ojs"},{"id":74,"value":"day2input\n .filter(([min, max, char, , str]) => {\n const pattern = new RegExp(`[^${char}]`, \"g\");\n const count = str.replaceAll(pattern, \"\").length;\n return min <= count && count <= max;\n }).length","pinned":true,"hidden":false,"mode":"ojs"},{"id":64,"value":"////////////////////\n// Part 2: given that same list of items, find the items which have the character in one of the two locations","pinned":true,"hidden":false,"mode":"ojs"},{"id":76,"value":"day2input.filter(([first, second, char, , str]) => {\n const [a, b] = [str[+first - 1], str[+second - 1]];\n return (a === char || b === char) && a !== b;\n}).length","pinned":true,"hidden":false,"mode":"ojs"},{"id":100,"value":"md`## Day 3`","pinned":false,"hidden":false,"mode":"ojs"},{"id":103,"value":"day3input = {\n const input = await FileAttachment(\"day 3.txt\").text();\n return input.split(\"\\n\").filter(line => line !== \"\");\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":108,"value":"// part 1 - how many times do you hit a hash sign if you start at the top left and move right 3, down 1 (over and over)","pinned":true,"hidden":false,"mode":"ojs"},{"id":110,"value":"function* slope(dx, dy) {\n let x = 0,\n y = 0;\n\n while (y < day3input.length) {\n yield day3input[y][x] === \"#\";\n x = (x + dx) % day3input[0].length;\n y += dy;\n }\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":147,"value":"[...slope(3, 1)]","pinned":true,"hidden":false,"mode":"ojs"},{"id":123,"value":"[...slope(3, 1)].filter(yn => yn === true).length","pinned":true,"hidden":false,"mode":"ojs"},{"id":134,"value":"// part 2","pinned":true,"hidden":false,"mode":"ojs"},{"id":135,"value":"md` given these slopes, multiply the number of trees you hit in each\n\n1. Right 1, down 1.\n1. Right 3, down 1. (This is the slope you already checked.)\n1. Right 5, down 1.\n1. Right 7, down 1.\n1. Right 1, down 2.\n`","pinned":false,"hidden":false,"mode":"ojs"},{"id":139,"value":"{\n const a = [...slope(1, 1)].filter(yn => yn === true).length;\n const b = [...slope(3, 1)].filter(yn => yn === true).length;\n const c = [...slope(5, 1)].filter(yn => yn === true).length;\n const d = [...slope(7, 1)].filter(yn => yn === true).length;\n const e = [...slope(1, 2)].filter(yn => yn === true).length;\n return a*b*c*d*e\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":268,"value":"md`# Day 4`\n","pinned":false,"hidden":false,"mode":"ojs"},{"id":177,"value":"day4input = FileAttachment(\"day 4.txt\")","pinned":false,"hidden":false,"mode":"ojs"},{"id":180,"value":"day4asObjects = (await day4input.text())\n .split(\n `\n\n`\n )\n .map(record =>\n Object.fromEntries(\n record\n .replaceAll(\"\\n\", \" \")\n .split(/ /)\n .map(i => i.split(\":\"))\n )\n )","pinned":true,"hidden":false,"mode":"ojs"},{"id":270,"value":"// Part 1: given an input where records are separated by blank lines, how many have the required fields","pinned":true,"hidden":false,"mode":"ojs"},{"id":168,"value":"$31","pinned":false,"hidden":false,"mode":"ojs"},{"id":208,"value":"day4asObjects.filter(record =>\n [\n { id: \"byr\" },\n { id: \"iyr\" },\n { id: \"eyr\" },\n { id: \"hgt\" },\n { id: \"hcl\" },\n { id: \"ecl\" },\n { id: \"pid\" }\n ].every(field => field.id in record)\n)","pinned":false,"hidden":false,"mode":"ojs"},{"id":263,"value":"// Part 2: validate specifics of the field value","pinned":true,"hidden":false,"mode":"ojs"},{"id":196,"value":"$32","pinned":false,"hidden":false,"mode":"ojs"},{"id":210,"value":"day4asObjects.filter(record =>\n [\n { id: \"byr\", test: v => /^\\d{4}$/.test(v) && 1920 <= v && v <= 2002 },\n { id: \"iyr\", test: v => /^\\d{4}$/.test(v) && 2010 <= v && v <= 2020 },\n { id: \"eyr\", test: v => /^\\d{4}$/.test(v) && 2020 <= v && v <= 2030 },\n {\n id: \"hgt\",\n test: v => {\n const n = v.slice(0, -2);\n return (\n (v.endsWith(\"cm\") && 150 <= n && n <= 193) ||\n (v.endsWith(\"in\") && 59 <= n && n <= 76)\n );\n }\n },\n { id: \"hcl\", test: v => /^#[0-9a-f]{6}$/.test(v) },\n {\n id: \"ecl\",\n test: v => [\"amb\", \"blu\", \"brn\", \"gry\", \"grn\", \"hzl\", \"oth\"].includes(v)\n },\n { id: \"pid\", test: v => /^\\d{9}$/.test(v) }\n ].every(({ id, test }) => id in record && test(record[id]))\n)","pinned":false,"hidden":false,"mode":"ojs"},{"id":292,"value":"md`## Day 5`","pinned":false,"hidden":false,"mode":"ojs"},{"id":313,"value":"day5inputFile = FileAttachment(\"day 5 input.txt\")","pinned":true,"hidden":false,"mode":"ojs"},{"id":329,"value":"day5input = (await day5inputFile.text()).split(\"\\n\").filter(e => e !== \"\")","pinned":false,"hidden":false,"mode":"ojs"},{"id":317,"value":"$33","pinned":false,"hidden":false,"mode":"ojs"},{"id":294,"value":"// Part 1","pinned":true,"hidden":false,"mode":"ojs"},{"id":296,"value":"$34","pinned":false,"hidden":false,"mode":"ojs"},{"id":324,"value":"// get the maximum ID for a boarding pass on the plane\nMath.max(...day5input.map(pass => day5BoardingPass(pass).id))","pinned":true,"hidden":false,"mode":"ojs"},{"id":298,"value":"// Part 2","pinned":true,"hidden":false,"mode":"ojs"},{"id":388,"value":"// find the ID# of your seat\n// (lazy hack: sort so that you can find the gap... the result here is one less than the id# of your seat)\n{\n const passengers = day5input.map(pass => day5BoardingPass(pass));\n passengers.sort(({ id: a }, { id: b }) => (a < b ? -1 : 1));\n\n return passengers\n .map((e, i, arr) => ({\n ...e,\n diff: arr[i + 1] ? arr[i + 1].id - e.id : \"end\"\n }))\n .find(e => e.diff !== 1);\n}","pinned":false,"hidden":false,"mode":"ojs"},{"id":291,"value":"md`

Part 2

`","pinned":false,"hidden":false,"mode":"ojs"},{"id":409,"value":"md`## Day 6`","pinned":true,"hidden":false,"mode":"ojs"},{"id":413,"value":"$35","pinned":false,"hidden":false,"mode":"ojs"},{"id":439,"value":"day6input = {\n return splitOnBlankLines(\n (await FileAttachment(\"day 6 input.txt\").text()).trim()\n );\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":411,"value":"// Part 1","pinned":true,"hidden":false,"mode":"ojs"},{"id":458,"value":"// determine how many questions each group answered \"yes\" to and sum them all\nday6input\n .map(e =>\n e\n // remove newlines\n .replaceAll(\"\\n\", \"\")\n // convert into an array\n .split(\"\")\n // dedupe - removes items which aren't the first index of the element\n .filter((e, i, arr) => arr.indexOf(e) === i)\n )\n // get length of each item (eg. how many unique answers)\n .map(arr => arr.length)\n // sum\n .reduce((acc, el) => acc + el, 0)","pinned":true,"hidden":false,"mode":"ojs"},{"id":415,"value":"// Part 2 - For each group, count the number of questions to which everyone answered \"yes\". What is the sum of those counts?","pinned":true,"hidden":false,"mode":"ojs"},{"id":479,"value":"day6input\n .map(e =>\n e\n // replace newlines with empty char\n .replaceAll(\"\\n\", \" \")\n // split on empty char to turn into array of persons\n .split(\" \")\n // find the entries which exist in every element of the array\n .reduce((acc, el, i, arr) => {\n if (typeof acc === \"string\") {\n // for the first iteration, the acc will be a string\n acc = acc.split(\"\");\n }\n\n // intersect the accumulator and current element\n return el.split(\"\").filter(item => acc.includes(item));\n })\n )\n .map(arr => arr.length)\n .reduce((a, e) => a + e, 0)","pinned":true,"hidden":false,"mode":"ojs"},{"id":523,"value":"md`## Day 7`","pinned":true,"hidden":false,"mode":"ojs"},{"id":534,"value":"day7input = (await FileAttachment(\"day 7 input.txt\").text()).trim().split(\"\\n\")","pinned":false,"hidden":false,"mode":"ojs"},{"id":539,"value":"day7rules = Object.fromEntries(\n day7input.map(e => {\n const [key, values] = e.split(\"contain\");\n\n return [\n key.replace(\" bags \", \"\"),\n Object.fromEntries(\n values\n .replaceAll(/ bags?\\.?/g, \"\")\n .split(\",\")\n .map(e => {\n const [, count, id] = e.trim().split(/(\\d) /);\n return [id, +count];\n })\n )\n ];\n })\n)","pinned":true,"hidden":false,"mode":"ojs"},{"id":288,"value":"$36","pinned":false,"hidden":false,"mode":"ojs"},{"id":569,"value":"function day7FindBagsContaining(bagTypes) {\n let allContainers = [];\n\n while (bagTypes.length > 0) {\n bagTypes = Object.entries(day7rules)\n // omit items we've checked already\n // .filter(([parent, children]) => !allContainers.includes(parent))\n // filter to the next level of parents\n .filter(el => {\n const [parent, children] = el;\n return Object.keys(children).some(child => bagTypes.includes(child));\n })\n .map(e => e[0]);\n\n allContainers = allContainers.concat(bagTypes);\n }\n\n return allContainers;\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":526,"value":"// Part 1 - How many bag colors can eventually contain at least one shiny gold bag? (The list of rules is quite long; make sure you get all of it.)","pinned":true,"hidden":false,"mode":"ojs"},{"id":573,"value":"day7FindBagsContaining([\"shiny gold\"])","pinned":true,"hidden":false,"mode":"ojs"},{"id":529,"value":"// Part 2","pinned":true,"hidden":false,"mode":"ojs"},{"id":619,"value":"function day7countBagsInside(bagTypes) {\n let count = 0;\n\n while (bagTypes.length > 0) {\n // get the item at the front of the bagtype list\n const type = bagTypes.shift();\n if (type !== \"undefined\") {\n // add the number of bags to the total\n count += Object.values(day7rules[type]).reduce((acc, el) => {\n // TODO: figure out how to count things\n return acc + (el ? el : 0);\n }, 0);\n\n // add the children bags onto the queue\n bagTypes.unshift(\n ...Object.keys(day7rules[type]).filter(type => type !== \"undefined\")\n );\n }\n }\n return count;\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":618,"value":"day7countBagsInside([\"shiny gold\"])","pinned":true,"hidden":false,"mode":"ojs"},{"id":433,"value":"md`---`","pinned":false,"hidden":false,"mode":"ojs"},{"id":522,"value":"// TEMPLATE:md`## Day 5`md`

Prompt

`// Part 1// Part 2","pinned":true,"hidden":false,"mode":"ojs"},{"id":447,"value":"md`## utils`","pinned":true,"hidden":false,"mode":"ojs"},{"id":449,"value":"function splitOnBlankLines(string) {\n return string.split(\n `\n\n`\n );\n}","pinned":true,"hidden":false,"mode":"ojs"},{"id":25,"value":"md`## Appendix`","pinned":false,"hidden":false,"mode":"ojs"},{"id":5,"value":"import {styles} from \"@sgregson/styles\"","pinned":false,"hidden":false,"mode":"ojs"},{"id":83,"value":"import { toc } from \"@sgregson/toc\"","pinned":false,"hidden":false,"mode":"ojs"},{"id":240,"value":"import { mdPlus } from \"@sgregson/md-plus\"","pinned":false,"hidden":false,"mode":"ojs"},{"id":7,"value":"styles","pinned":false,"hidden":false,"mode":"ojs"},{"id":10,"value":"title = (day) => md`# 2020 (Advent of Code)${(day && ` - Day ${day}`) || \"\"}`","pinned":false,"hidden":false,"mode":"ojs"},{"id":342,"value":"import { Debug } from \"@sgregson/utils\"","pinned":false,"hidden":false,"mode":"ojs"}],"files":"$W37","stdlib":"1","titling":"auto"}