Merge branch 'master' of ssh://git.elnota.space:10222/daviddoji/Advent_of_code

src/Year_2015/P8.py

@@ -0,0 +1,83 @@
+# --- Day 8: Matchsticks ---
+
+# Space on the sleigh is limited this year, and so Santa will be bringing his
+# list as a digital copy. He needs to know how much space it will take up when
+# stored.
+
+# It is common in many programming languages to provide a way to escape special
+# characters in strings. For example, C, JavaScript, Perl, Python, and even PHP
+# handle special characters in very similar ways.
+
+# However, it is important to realize the difference between the number of
+# characters in the code representation of the string literal and the number of
+# characters in the in-memory string itself.
+
+# For example:
+
+#     "" is 2 characters of code (the two double quotes), but the string
+# contains zero characters.
+#     "abc" is 5 characters of code, but 3 characters in the string data.
+#     "aaa\"aaa" is 10 characters of code, but the string itself contains six
+# "a" characters and a single, escaped quote character, for a total of 7 characters in the string data.
+#     "\x27" is 6 characters of code, but the string itself contains just one
+# - an apostrophe ('), escaped using hexadecimal notation.
+
+# Santa's list is a file that contains many double-quoted string literals, one
+# on each line. The only escape sequences used are \\ (which represents a
+# single backslash), \" (which represents a lone double-quote character), and
+# \x plus two hexadecimal characters (which represents a single character with
+# that ASCII code).
+
+# Disregarding the whitespace in the file, what is the number of characters of
+# code for string literals minus the number of characters in memory for the
+# values of the strings in total for the entire file?
+
+# For example, given the four strings above, the total number of characters of
+# string code (2 + 5 + 10 + 6 = 23) minus the total number of characters in
+# memory for string values (0 + 3 + 7 + 1 = 11) is 23 - 11 = 12.
+
+with open("files/P8.txt") as f:
+    lines = [line for line in f.read().strip().split()]
+
+
+# TIL: eval()
+def part_1() -> None:
+    res = sum(len(line) - len(eval(line)) for line in lines)
+
+    print(f"There are {res} characters.")
+
+
+# --- Part Two ---
+
+# Now, let's go the other way. In addition to finding the number of characters
+# of code, you should now encode each code representation as a new string and
+# find the number of characters of the new encoded representation, including
+# the surrounding double quotes.
+
+# For example:
+
+#     "" encodes to "\"\"", an increase from 2 characters to 6.
+#     "abc" encodes to "\"abc\"", an increase from 5 characters to 9.
+#     "aaa\"aaa" encodes to "\"aaa\\\"aaa\"", an increase from 10 characters to
+# 16.
+#     "\x27" encodes to "\"\\x27\"", an increase from 6 characters to 11.
+
+# Your task is to find the total number of characters to represent the newly
+# encoded strings minus the number of characters of code in each original
+# string literal. For example, for the strings above, the total encoded length
+# (6 + 9 + 16 + 11 = 42) minus the characters in the original code
+# representation (23, just like in the first part of this puzzle) is
+# 42 - 23 = 19.
+
+
+def part_2() -> None:
+    # 2 for the ""
+    # just count numbers of \\ and new "
+    res = sum(2 + line.count("\\") + line.count('"') for line in lines)
+
+    print(f"There are {res} characters.")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2015/files/P8.txt

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src/Year_2016/P8.py

@@ -0,0 +1,133 @@
+# --- Day 8: Two-Factor Authentication ---
+
+# You come across a door implementing what you can only assume is an
+# implementation of two-factor authentication after a long game of
+# requirements telephone.
+
+# To get past the door, you first swipe a keycard (no problem; there was one on
+# a nearby desk). Then, it displays a code on a little screen, and you type
+# that code on a keypad. Then, presumably, the door unlocks.
+
+# Unfortunately, the screen has been smashed. After a few minutes, you've taken
+# everything apart and figured out how it works. Now you just have to work out
+# what the screen would have displayed.
+
+# The magnetic strip on the card you swiped encodes a series of instructions
+# for the screen; these instructions are your puzzle input. The screen is 50
+# pixels wide and 6 pixels tall, all of which start off, and is capable of
+# three somewhat peculiar operations:
+
+#     rect AxB turns on all of the pixels in a rectangle at the top-left of the
+# screen which is A wide and B tall.
+#     rotate row y=A by B shifts all of the pixels in row A (0 is the top row)
+# right by B pixels. Pixels that would fall off the right end appear at the
+# left end of the row.
+#     rotate column x=A by B shifts all of the pixels in column A (0 is the
+# left column) down by B pixels. Pixels that would fall off the bottom appear
+# at the top of the column.
+
+# For example, here is a simple sequence on a smaller screen:
+
+#     rect 3x2 creates a small rectangle in the top-left corner:
+
+#     ###....
+#     ###....
+#     .......
+
+#     rotate column x=1 by 1 rotates the second column down by one pixel:
+
+#     #.#....
+#     ###....
+#     .#.....
+
+#     rotate row y=0 by 4 rotates the top row right by four pixels:
+
+#     ....#.#
+#     ###....
+#     .#.....
+
+#     rotate column x=1 by 1 again rotates the second column down by one pixel,
+# causing the bottom pixel to wrap back to the top:
+
+#     .#..#.#
+#     #.#....
+#     .#.....
+
+# As you can see, this display technology is extremely powerful, and will soon
+# dominate the tiny-code-displaying-screen market. That's what the
+# advertisement on the back of the display tries to convince you, anyway.
+
+# There seems to be an intermediate check of the voltage used by the display:
+# after you swipe your card, if the screen did work, how many pixels should be
+# lit?
+
+import numpy as np
+import numpy.typing as npt
+
+with open("files/P8.txt") as f:
+    instructions = [line for line in f.read().strip().split("\n")]
+
+
+def rect(instruction: str, grid: npt.NDArray[int]) -> npt.NDArray[int]:
+    _, row = instruction.split("x")
+    _, col = _.split()
+    grid[: int(row), : int(col)] = 1
+
+
+def rotate_column(
+    instruction: str, grid: npt.NDArray[int]
+) -> npt.NDArray[int]:
+    _, cols = instruction.split("=")
+    x, _, steps = cols.split()
+    new_col = np.roll(grid[:, int(x) : int(x) + 1], int(steps), axis=0)
+    grid[:, int(x) : int(x) + 1] = new_col
+
+
+def rotate_row(instruction: str, grid: npt.NDArray[int]) -> npt.NDArray[int]:
+    _, rows = instruction.split("=")
+    y, _, steps = rows.split()
+    new_row = np.roll(grid[int(y) : int(y) + 1, :], int(steps), axis=1)
+    grid[int(y) : int(y) + 1, :] = new_row
+
+
+# dimensions given by the problem
+init_grid = np.zeros((6, 50), dtype=int)
+
+
+def part_1() -> None:
+
+    for instruction in instructions:
+        if instruction.startswith("rect"):
+            rect(instruction, init_grid)
+        elif instruction.startswith("rotate row"):
+            rotate_row(instruction, init_grid)
+        elif instruction.startswith("rotate column"):
+            rotate_column(instruction, init_grid)
+
+    lit = np.sum(init_grid)
+
+    print(f"There should be {lit} pixels lit")
+
+
+# --- Part Two ---
+
+# You notice that the screen is only capable of displaying capital letters; in
+# the font it uses, each letter is 5 pixels wide and 6 tall.
+
+# After you swipe your card, what code is the screen trying to display?
+
+
+def display(grid: npt.NDArray[int]) -> None:
+    print(
+        "\n".join("".join("X" if one else " " for one in row) for row in grid)
+    )
+
+
+def part_2() -> None:
+    print("The code is the following:\n")
+    display(init_grid)
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2016/files/P8.txt

@@ -0,0 +1,173 @@
+rect 1x1
+rotate row y=0 by 20
+rect 1x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 3
+rect 2x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 3
+rect 2x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 4
+rect 2x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 3
+rect 2x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 5
+rect 1x1
+rotate row y=0 by 2
+rect 1x1
+rotate row y=0 by 6
+rect 5x1
+rotate row y=0 by 2
+rect 1x3
+rotate row y=2 by 8
+rotate row y=0 by 8
+rotate column x=0 by 1
+rect 7x1
+rotate row y=2 by 24
+rotate row y=0 by 20
+rotate column x=5 by 1
+rotate column x=4 by 2
+rotate column x=2 by 2
+rotate column x=0 by 1
+rect 7x1
+rotate column x=34 by 2
+rotate column x=22 by 1
+rotate column x=15 by 1
+rotate row y=2 by 18
+rotate row y=0 by 12
+rotate column x=8 by 2
+rotate column x=7 by 1
+rotate column x=5 by 2
+rotate column x=2 by 1
+rotate column x=0 by 1
+rect 9x1
+rotate row y=3 by 28
+rotate row y=1 by 28
+rotate row y=0 by 20
+rotate column x=18 by 1
+rotate column x=15 by 1
+rotate column x=14 by 1
+rotate column x=13 by 1
+rotate column x=12 by 2
+rotate column x=10 by 3
+rotate column x=8 by 1
+rotate column x=7 by 2
+rotate column x=6 by 1
+rotate column x=5 by 1
+rotate column x=3 by 1
+rotate column x=2 by 2
+rotate column x=0 by 1
+rect 19x1
+rotate column x=34 by 2
+rotate column x=24 by 1
+rotate column x=23 by 1
+rotate column x=14 by 1
+rotate column x=9 by 2
+rotate column x=4 by 2
+rotate row y=3 by 5
+rotate row y=2 by 3
+rotate row y=1 by 7
+rotate row y=0 by 5
+rotate column x=0 by 2
+rect 3x2
+rotate column x=16 by 2
+rotate row y=3 by 27
+rotate row y=2 by 5
+rotate row y=0 by 20
+rotate column x=8 by 2
+rotate column x=7 by 1
+rotate column x=5 by 1
+rotate column x=3 by 3
+rotate column x=2 by 1
+rotate column x=1 by 2
+rotate column x=0 by 1
+rect 9x1
+rotate row y=4 by 42
+rotate row y=3 by 40
+rotate row y=1 by 30
+rotate row y=0 by 40
+rotate column x=37 by 2
+rotate column x=36 by 3
+rotate column x=35 by 1
+rotate column x=33 by 1
+rotate column x=32 by 1
+rotate column x=31 by 3
+rotate column x=30 by 1
+rotate column x=28 by 1
+rotate column x=27 by 1
+rotate column x=25 by 1
+rotate column x=23 by 3
+rotate column x=22 by 1
+rotate column x=21 by 1
+rotate column x=20 by 1
+rotate column x=18 by 1
+rotate column x=17 by 1
+rotate column x=16 by 3
+rotate column x=15 by 1
+rotate column x=13 by 1
+rotate column x=12 by 1
+rotate column x=11 by 2
+rotate column x=10 by 1
+rotate column x=8 by 1
+rotate column x=7 by 2
+rotate column x=5 by 1
+rotate column x=3 by 3
+rotate column x=2 by 1
+rotate column x=1 by 1
+rotate column x=0 by 1
+rect 39x1
+rotate column x=44 by 2
+rotate column x=42 by 2
+rotate column x=35 by 5
+rotate column x=34 by 2
+rotate column x=32 by 2
+rotate column x=29 by 2
+rotate column x=25 by 5
+rotate column x=24 by 2
+rotate column x=19 by 2
+rotate column x=15 by 4
+rotate column x=14 by 2
+rotate column x=12 by 3
+rotate column x=9 by 2
+rotate column x=5 by 5
+rotate column x=4 by 2
+rotate row y=5 by 5
+rotate row y=4 by 38
+rotate row y=3 by 10
+rotate row y=2 by 46
+rotate row y=1 by 10
+rotate column x=48 by 4
+rotate column x=47 by 3
+rotate column x=46 by 3
+rotate column x=45 by 1
+rotate column x=43 by 1
+rotate column x=37 by 5
+rotate column x=36 by 5
+rotate column x=35 by 4
+rotate column x=33 by 1
+rotate column x=32 by 5
+rotate column x=31 by 5
+rotate column x=28 by 5
+rotate column x=27 by 5
+rotate column x=26 by 3
+rotate column x=25 by 4
+rotate column x=23 by 1
+rotate column x=17 by 5
+rotate column x=16 by 5
+rotate column x=13 by 1
+rotate column x=12 by 5
+rotate column x=11 by 5
+rotate column x=3 by 1
+rotate column x=0 by 1

src/Year_2017/P7.py

@@ -88,5 +88,78 @@ def part_1() -> None:
     print(f"The name of the bottom program is {root_program}")
 
 
+# --- Part Two ---
+
+# The programs explain the situation: they can't get down. Rather, they could
+# get down, if they weren't expending all of their energy trying to keep the
+# tower balanced. Apparently, one program has the wrong weight, and until it's
+# fixed, they're stuck here.
+
+# For any program holding a disc, each program standing on that disc forms a
+# sub-tower. Each of those sub-towers are supposed to be the same weight, or
+# the disc itself isn't balanced. The weight of a tower is the sum of the
+# weights of the programs in that tower.
+
+# In the example above, this means that for ugml's disc to be balanced, gyxo,
+# ebii, and jptl must all have the same weight, and they do: 61.
+
+# However, for tknk to be balanced, each of the programs standing on its disc
+# and all programs above it must each match. This means that the following sums
+# must all be the same:
+
+#     ugml + (gyxo + ebii + jptl) = 68 + (61 + 61 + 61) = 251
+#     padx + (pbga + havc + qoyq) = 45 + (66 + 66 + 66) = 243
+#     fwft + (ktlj + cntj + xhth) = 72 + (57 + 57 + 57) = 243
+
+# As you can see, tknk's disc is unbalanced: ugml's stack is heavier than the
+# other two. Even though the nodes above ugml are balanced, ugml itself is too
+# heavy: it needs to be 8 units lighter for its stack to weigh 243 and keep the
+# towers balanced. If this change were made, its weight would be 60.
+
+# Given that exactly one program is the wrong weight, what would its weight
+# need to be to balance the entire tower?
+
+
+def child_values(node, node_map, node_values):
+    weights, unbalanced = [], []
+    children = node_map[node]
+    for child in children:
+        if child in node_map.keys():
+            child_weight, child_balance = child_values(
+                child, node_map, node_values
+            )
+            value = sum(child_weight) + node_values[child]
+            unbalanced.append(child_balance)
+        else:
+            value = node_values[child]
+        weights.append(value)
+    if len(set(weights)) != 1:
+        unbalanced.append((node_map[node], weights))
+    return weights, unbalanced
+
+
+def part_2() -> None:
+    node_map, node_values = {}, {}
+    for line in towers:
+        if "->" in line:
+            lhs, rhs = line.split(" -> ")
+            parent, value = lhs.split()
+            node_values[parent] = int(value[1:-1])
+            _childs = rhs.split(", ")
+            node_map[parent] = [child for child in _childs]
+        else:
+            child, value = line.split()
+            node_values[child] = int(value[1:-1])
+
+    _, unbalance = child_values("rqwgj", node_map, node_values)
+
+    heavier_child_idx = unbalance[0][0][5][1].index(max(unbalance[0][0][5][1]))
+    heavier_child = unbalance[0][0][5][0][heavier_child_idx]
+    unbalanced_child = node_values[heavier_child]
+
+    print(f"To balance the tower, the weight must be {unbalanced_child - 8}")
+
+
 if __name__ == "__main__":
     part_1()
+    part_2()

src/Year_2021/P7.py

@@ -0,0 +1,117 @@
+# --- Day 7: The Treachery of Whales ---
+
+# A giant whale has decided your submarine is its next meal, and it's much
+# faster than you are. There's nowhere to run!
+
+# Suddenly, a swarm of crabs (each in its own tiny submarine - it's too deep
+# for them otherwise) zooms in to rescue you! They seem to be preparing to
+# blast a hole in the ocean floor; sensors indicate a massive underground cave
+# system just beyond where they're aiming!
+
+# The crab submarines all need to be aligned before they'll have enough power
+# to blast a large enough hole for your submarine to get through. However, it
+# doesn't look like they'll be aligned before the whale catches you! Maybe you
+# can help?
+
+# There's one major catch - crab submarines can only move horizontally.
+
+# You quickly make a list of the horizontal position of each crab (your puzzle
+# input). Crab submarines have limited fuel, so you need to find a way to make
+# all of their horizontal positions match while requiring them to spend as
+# little fuel as possible.
+
+# For example, consider the following horizontal positions:
+
+# 16,1,2,0,4,2,7,1,2,14
+
+# This means there's a crab with horizontal position 16, a crab with horizontal
+# position 1, and so on.
+
+# Each change of 1 step in horizontal position of a single crab costs 1 fuel.
+# You could choose any horizontal position to align them all on, but the one
+# that costs the least fuel is horizontal position 2:
+
+#     Move from 16 to 2: 14 fuel
+#     Move from 1 to 2: 1 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 0 to 2: 2 fuel
+#     Move from 4 to 2: 2 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 7 to 2: 5 fuel
+#     Move from 1 to 2: 1 fuel
+#     Move from 2 to 2: 0 fuel
+#     Move from 14 to 2: 12 fuel
+
+# This costs a total of 37 fuel. This is the cheapest possible outcome; more
+# expensive outcomes include aligning at position 1 (41 fuel), position 3 (39
+# fuel), or position 10 (71 fuel).
+
+# Determine the horizontal position that the crabs can align to using the least
+# fuel possible. How much fuel must they spend to align to that position?
+
+with open("files/P7.txt") as f:
+    positions = [int(num) for num in f.read().strip().split(",")]
+
+
+def part_1() -> None:
+    fuel = []
+    for pos in positions:
+        sum = 0
+        for elem in positions:
+            sum += abs(elem - pos)
+        fuel.append(sum)
+
+    print(f"It is needed {min(fuel)} units of fuel")
+
+
+# --- Part Two ---
+
+# The crabs don't seem interested in your proposed solution. Perhaps you
+# misunderstand crab engineering?
+
+# As it turns out, crab submarine engines don't burn fuel at a constant rate.
+# Instead, each change of 1 step in horizontal position costs 1 more unit of
+# fuel than the last: the first step costs 1, the second step costs 2, the
+# third step costs 3, and so on.
+
+# As each crab moves, moving further becomes more expensive. This changes the
+# best horizontal position to align them all on; in the example above, this
+# becomes 5:
+
+#     Move from 16 to 5: 66 fuel
+#     Move from 1 to 5: 10 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 0 to 5: 15 fuel
+#     Move from 4 to 5: 1 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 7 to 5: 3 fuel
+#     Move from 1 to 5: 10 fuel
+#     Move from 2 to 5: 6 fuel
+#     Move from 14 to 5: 45 fuel
+
+# This costs a total of 168 fuel. This is the new cheapest possible outcome;
+# the old alignment position (2) now costs 206 fuel instead.
+
+# Determine the horizontal position that the crabs can align to using the least
+# fuel possible so they can make you an escape route! How much fuel must they
+# spend to align to that position?
+
+# TIL :D
+def triangular_cost(steps: int) -> int:
+    return steps * (steps + 1) // 2
+
+
+def part_2() -> None:
+    fuel = []
+    for pos in positions:
+        sum = 0
+        for elem in positions:
+            sum += triangular_cost(abs(elem - pos))
+        fuel.append(sum)
+
+    print(f"It is needed {min(fuel)} units of fuel")
+
+
+if __name__ == "__main__":
+    part_1()
+    part_2()

src/Year_2021/files/P7.txt

@@ -0,0 +1 @@
+1101,1,29,67,1102,0,1,65,1008,65,35,66,1005,66,28,1,67,65,20,4,0,1001,65,1,65,1106,0,8,99,35,67,101,99,105,32,110,39,101,115,116,32,112,97,115,32,117,110,101,32,105,110,116,99,111,100,101,32,112,114,111,103,114,97,109,10,20,1091,861,228,628,186,980,996,710,541,354,1611,69,1331,91,1220,447,523,38,1286,244,643,1069,566,70,155,1710,1266,120,302,72,232,387,1086,278,1122,605,1559,98,111,1816,795,543,1217,304,356,129,839,704,49,523,370,74,13,232,179,101,664,892,266,622,197,404,147,882,435,504,48,766,684,1362,136,830,1393,1259,925,68,879,251,43,1339,61,98,403,51,1008,197,659,195,1823,233,121,731,82,141,580,18,427,774,13,685,496,752,63,132,39,237,18,167,51,299,22,19,1442,305,1283,253,159,731,302,76,115,185,136,447,821,307,207,30,1427,251,589,0,1096,1240,261,442,757,5,172,847,858,382,425,79,402,166,1058,186,35,21,324,183,1293,95,410,321,12,155,88,409,40,428,1180,199,1444,487,148,57,1187,15,100,983,77,0,1667,359,513,659,339,142,968,994,787,0,443,183,133,538,4,332,1459,204,1156,710,1654,20,36,407,890,1265,1090,743,36,78,1033,781,608,476,103,1401,24,4,875,414,799,305,1047,842,72,497,362,270,73,12,9,0,21,11,51,1357,455,505,483,552,199,1108,214,238,686,1496,116,154,1403,35,272,738,1024,50,50,934,564,19,395,324,447,794,1326,14,407,1694,452,439,455,442,86,1515,588,809,224,112,156,21,1405,610,187,23,370,112,397,995,777,75,1281,32,60,284,388,916,555,200,675,20,320,32,398,104,113,447,113,351,396,322,36,1674,1708,232,1004,514,95,859,382,116,277,239,343,3,433,11,55,699,513,1465,319,44,306,224,615,482,695,421,300,321,283,579,323,102,275,17,723,632,713,277,801,222,130,189,1549,49,784,690,136,444,1315,259,1334,472,711,1109,276,70,315,838,35,328,766,1100,460,4,178,630,571,5,106,429,368,547,1210,840,162,166,10,403,880,287,44,1001,316,402,1054,174,7,1194,105,58,268,667,86,588,166,547,238,1586,77,112,244,107,63,873,1152,16,407,198,26,587,20,1449,775,653,1369,732,262,566,222,10,102,22,573,233,297,1238,789,291,93,206,1274,177,58,841,672,37,16,262,201,241,938,133,774,978,631,511,0,263,498,799,51,330,11,206,325,173,676,15,457,364,46,373,34,1475,530,672,295,55,4,297,274,1519,15,688,555,96,160,185,583,646,41,378,1572,67,219,572,143,16,286,65,788,886,243,883,1012,109,90,742,464,1099,388,1855,731,62,6,415,66,232,542,98,1123,11,414,1262,4,440,8,691,130,164,773,992,423,115,1807,1618,153,168,1213,719,291,316,311,110,24,608,0,127,131,142,196,232,75,248,412,275,1295,239,86,967,133,422,415,894,280,807,345,446,250,979,231,713,201,1009,208,444,43,1400,434,8,221,141,235,909,1018,340,0,178,1144,353,662,491,294,6,440,446,824,1392,379,269,1427,911,671,231,424,102,718,86,54,130,206,514,137,1075,1573,248,472,602,249,974,8,372,59,2,940,165,1132,327,1424,63,6,97,140,1302,439,1237,59,324,733,397,477,426,278,274,1636,745,41,269,257,51,173,503,88,1223,754,228,584,72,632,645,323,156,7,337,192,375,583,613,370,172,528,1282,360,208,17,208,802,22,67,290,242,458,809,6,1388,49,14,327,911,121,475,101,617,284,91,1,35,421,293,1419,602,143,142,168,657,472,219,345,411,115,387,5,494,383,348,85,1070,154,42,63,586,953,563,12,1263,788,762,222,351,730,42,643,877,522,775,698,564,604,155,191,430,5,386,787,120,470,433,564,955,411,1090,210,1096,933,109,70,279,287,736,390,1075,30,194,30,1318,473,727,100,584,1227,89,432,120,393,1080,185,500,847,117,319,54,160,32,114,700,12,36,681,53,1063,113,82,1625,1426,53,658,80,253,36,16,810,560,602,177,275,147,335,1237,1447,176,55,366,721,471,501,217,67,899,914,168,981,1177,898,1,9,704,602,787,1280,305,57,786,1696,211,63,55,351,14,151,932,757,810,1748,169,1018,125,849,234,102,823,300,127,6,849,1163,229,726,397,656,135,466,137,1247,811,807,366,209,1703,24,1219,45,161,353,274,572,397,899,646,32,137,439,1048,2,391,8,214,736,518,409,414,567,1262,155,102,178,1247,526,10,94,759,781,18,14,1518,68,295,905,478,1581,89,429,937,438,915,1110,659,615,1128,8,125,89,289,1084,905,254,1227,184,883,983,110,1,24,748,1408,828,1187,63,264,481,214