Draw this Ascii coffee cup:

  o
o
o
__________
/          \__
|   J      |  \
|    A     |  |
|     V    |  |
|      A   |__/
\__________/

Brownie points for coffee-script or java :)

Shortest code in bytes, function or program, trailing newline or white space is acceptable, drink up!

• I'd be very suspicious of a cup of sparkling coffee. ;) Feb 24, 2017 at 14:22
• @Dennis it's my special coffee for Friday mornings ;) Feb 24, 2017 at 14:22
• Wouldn't be this more interesting with 2 or more drinks: the hot one would have vapors symbolized with “(” and “)”, the cold one sparkles? And by borrowing from Rod's comment, the code should display one or other based on current time. Feb 24, 2017 at 14:31
• Would it be acceptable to have trailing white space on lines? Feb 24, 2017 at 14:38
• @Aaron the cup don't have a good pattern, hard-coding/compressing will be shorter in many languages
– Rod
Feb 24, 2017 at 14:55

## JavaScript (ES6), 110 104 bytes

Saved 4 bytes thanks to edc65

let f =

_=>1o
6o
3o
9
/44\\__
|2J5|1\\
|3A4|1|
|4V3|1|
|5A2|__/
\\9/.replace(/\d/g,n=>' _'[n>>3].repeat(++n))

console.log(f())

### How it works

The compression of the original ASCII art is achieved by replacing all sequences of 2 to 10 consecutive spaces and the two sequences of 10 consecutive underscores with a single digit:

• Each sequence of N consecutive spaces is encoded with the digit N-1.
• The underscore sequences are encoded with a 9.

We use N-1 rather than N so that we never have to use more than one digit. Hence the need for ++n when decoding.

The expression n>>3 (bitwise shift to the right) equals 0 for n = 1 to n = 7 and equals 1 for n = 8 (not used) and n = 9. Therefore, ' _'[n>>3] gives an underscore for 9, and a space for all other encountered values.

The only special case is the sequence of 10 consecutive spaces just above "JAVA". Encoding it with a 9 would conflict with the underscore sequences. So we need to split it into two sequences of 5 spaces, encoded as 44.

• I count 108 bytes (not counting f=). You can save 4 bytes this way: n>>3 instead of +!(n&7), 9 instead of _8 (twice) and 44 instead of 9 Feb 24, 2017 at 15:19
• @edc65 I've no idea why I counted f= in that one... Thanks for the saved bytes! Feb 24, 2017 at 15:28
• Can you explain how the regex works a little bit? It appears like it replaced a digit d, with d spaces ('4' becomes ' '). But not sure exactly how it does that. What does the bit shift do? Why are we incrementing n? Feb 24, 2017 at 16:33
• @Cruncher I've added a 'How it works' section. Feb 24, 2017 at 17:22
• @Arnauld Very clever :) Feb 24, 2017 at 17:43

# Jelly, 67 64 bytes

-2 bytes thanks to Dennis (1. remove redundant ”, and 2. replace transpose and run-length decode, ZŒṙ, with reduce by element repetition, x/.)

“Ñṁ{xGgṭḷVỤɲ8ṿfƬT9Ɱ¹=qṀS“$<(ƇỤ08ØÑḌṃṘX6~cuc8HṗḞ2’Dx/ị“ ¶_/\|JAVo  Try it online! ### How? “...“...’ is a list of two base-250 compressed numbers: [1021021021332411532617161526181616261916162618163425334, 2117114111551155121131612111415121115141211161312111551]  D converts to decimal to yield two lists of digits: [[1, 0, 2, 1, 0, 2, 1, 0, 2, 1, 3, 3, 2, 4, 1, 1, 5, 3, 2, 6, 1, 7, 1, 6, 1, 5, 2, 6, 1, 8, 1, 6, 1, 6, 2, 6, 1, 9, 1, 6, 1, 6, 2, 6, 1, 8, 1, 6, 3, 4, 2, 5, 3, 3, 4], [2, 1, 1, 7, 1, 1, 4, 1, 1, 1, 5, 5, 1, 1, 5, 5, 1, 2, 1, 1, 3, 1, 6, 1, 2, 1, 1, 1, 4, 1, 5, 1, 2, 1, 1, 1, 5, 1, 4, 1, 2, 1, 1, 1, 6, 1, 3, 1, 2, 1, 1, 1, 5, 5, 1]]  x/ reduces by element repetition to give one list of digits (repeating the number from the first list by the corresponding value of the other): [1, 1, 0, 2, 1, 1, 1, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 0, 2, 1, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 2, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 3, 3, 2, 6, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 6, 1, 1, 5, 2, 6, 1, 1, 1, 1, 8, 1, 1, 1, 1, 1, 6, 1, 1, 6, 2, 6, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 6, 1, 1, 6, 2, 6, 1, 1, 1, 1, 1, 1, 8, 1, 1, 1, 6, 3, 3, 4, 2, 5, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4]  ị instructs to index into the list of the right, one based and modularly (0 indexes into the rightmost item). The list on the right,  ¶_/\|JAVo, is simply the character used in the required order where the pilcrow, ¶, is the same code-point as a linefeed. The closing partner of “ is not required as this is the end of the program: [' ', ' ', 'o', '\n', ' ', ' ', ' ', ' ', ' ', ' ', ' ', 'o', '\n', ' ', ' ', ' ', ' ', 'o', '\n', ' ', '_', '_', '_', '_', '_', '_', '_', '_', '_', '_', '\n', '/', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', ' ', '\\', '_', '_', '\n', '|', ' ', ' ', ' ', 'J', ' ', ' ', ' ', ' ', ' ', ' ', '|', ' ', ' ', '\\', '\n', '|', ' ', ' ', ' ', ' ', 'A', ' ', ' ', ' ', ' ', ' ', '|', ' ', ' ', '|', '\n', '|', ' ', ' ', ' ', ' ', ' ', 'V', ' ', ' ', ' ', ' ', '|', ' ', ' ', '|', '\n', '|', ' ', ' ', ' ', ' ', ' ', ' ', 'A', ' ', ' ', ' ', '|', '_', '_', '/', '\n', '\\', '_', '_', '_', '_', '_', '_', '_', '_', '_', '_', '/']  Jelly performs an implicit print of this list, which, since it contains characters, prints as if it were a string:  o o o __________ / \__ | J | \ | A | | | V | | | A |__/ \__________/  • I swear some of these languages are straight up compression algorithms Feb 24, 2017 at 16:26 • @Cruncher that would be Bubblegum Feb 24, 2017 at 16:32 • Of course, any language that can output text longer than the code, must necessarily have code longer than the output text for some output texts. I assume if you tried to write something for a completely randomized input the code for it(unless you got lucky) would be longer? Feb 24, 2017 at 16:38 • Yep if random. Bubblegum is actually using compression, the goal being Kolmogorov complexity challenges and hence the input should have pattern (or at least repetition like here). Feb 24, 2017 at 16:51 • The final ” is implicit and you can replace ZŒṙ with x/. Also, while it doesn't have any bytes, using ¶ instead of a literal newline makes the code more redable imo. Feb 26, 2017 at 13:43 # CoffeeScript ES6, 214 180 bytes r="replace";" 1o0n0 6o0n0 3o0n0 _9n0/0 9b0_1n0|0 2J0 5|0 1b0n0|0 3A 4|0 1|0n0|0 4V0 3|0 1|0n0|0 5A0 2|0_1/0n0b0_9/0"[r](/\d/g,(a,b,c)->c[b-1].repeat(a))[r](/n/g,"\n")[r](/b/g,"\\")  # CoffeeScript, 135 bytes with hardcoding f=()->""" o o o __________ / \__ | J | \\ | A | | | V | | | A |__/ \__________/"""  • No up/down vote; I don't like this answer because generally the point in a kolmogorov-complexity answer is to generate the output without using the entire thing in the code. Feb 24, 2017 at 14:37 • @HyperNeutrino, I agree, working on improving it. – Tom Feb 24, 2017 at 14:39 # PowerShell, 136124123 105 bytes """2o 7o 4o$(($a='_'*10)) /55\__ |3J6|2\ |4A5|2| |5V4|2| |6A3|__/ \$a/"""-replace'(\d)','$(" "*$1)'|iex


Try it online!

Thanks to @briantist for finding the shorter -replace method that I knew was there somewhere.

This takes the string with numbers in place of the requisite number of spaces. We then regex -replace the digits with a script expression $(" "*$1). So, for example, the first line of the string will be $(" "*2)o, the second will be $(" "*7)o and so on. Because of the triple-quoting, this is left as a string on the pipeline. We dump that to iex (short for Invoke-Expression and similar to eval), which processes the script expressions and leaves the resulting multi-line string on the pipeline. Output is implicit.

• That's strange, hardcoding is shorter. Hm. +1 anyway :) Feb 24, 2017 at 14:47
• I was hoping for some answers using innovative (huffman) coding schemas, but my python implementation is still coming up longer as well.. Feb 24, 2017 at 14:54
• for some reason the short-hand if/else doesn't seem to work ($_,' '*$_)[+$_-in48..57] - no matter what I change it seems to fail for me. Feb 24, 2017 at 17:43 • @ConnorLSW That's because both expressions are evaluated and the array built before the indexing happens. As a result, PowerShell doesn't know how to multiply space by o and barfs. Feb 24, 2017 at 17:54 • @briantist Thanks for finding that! I knew it was there, just couldn't come up with the right combination of quotes to get it to work. Feb 27, 2017 at 13:47 # Python 2, 174172171 167 bytes No hard-coding. No Base-64 encoding. No Regex. k=' ' q='_'*10 print'\n'.join([k*i+'o'for i in 2,7,4]+[k+q]+['/'+k*10+'\\__']+['|'+k*s+'JAVA'[s-3]+k*(9-s)+'|'+' _'[s==6]*2+'\\||/'[s-3]for s in 3,4,5,6]+['\\'+q+'/'])  Saved 2 bytes by externalizing '_'*10 and by exploiting Python's conversion of True -> 1 and False -> 0. Saved 1 byte by removing unnecessary whitespace. Saved 4 bytes thanks to @TuukkaX! • You seem to have 2 useless whitespaces at ] for and in [. Feb 25, 2017 at 7:07 • Actually, you can shorten [2,7,4] and [3,4,5,6] to 2,4,7 and 3,4,5,6. Feb 25, 2017 at 7:09 ## GNU sed, 113 112 bytes s:$:  o@SS o@S o@ UU@/SSS \\__@|SJSS|  \\@|S AS  |  |@|S  VS |  |@|SSAS|__/@\\UU/:
s:S:   :g
y:@:\n:
s:U:_____:g


Basic encoding, it stores 3 spaces as S, \n as @ and 5 underlines as U. I'll keep trying combinations to find something shorter.

Try it online!

The trivial solution of printing the string directly is given below. It has 136 bytes, resulting in a compression of 18 %, using the encoding scheme above.

c\  o\
o\
o\
__________\
/          \\__\
|   J      |  \\\
|    A     |  |\
|     V    |  |\
|      A   |__/\
\\__________/


Try it online!

• Encoding ss as S saves 1 byte. Feb 24, 2017 at 16:19
• @Riley Thanks. I also just found a 1 byte less solution, with S storing 3 spaces, no s. I think I'll edit this one instead, because it keeps the same number of transformations. Feb 24, 2017 at 16:26

# MATL, 87868382 78 bytes

[TIH][IAC]111Z?c'(ty!(OWM4J4gW{lm> >bw8ch|.FU2W"@\#2Dj!NQDeIMZ'F'_ /|\JAV'Za7e


This solution breaks the coffee into two pieces: the "bubbles" and the mug. To create the bubbles, we create a sparse matrix with 111 located at three locations and convert it to a character array

[TIH][IAC]111Z?c


For the mug component, we rely upon string compression

'(ty!(OWM4J4gW{lm> >bw8ch|.FU2W"@\#2Dj!NQDeIMZ'F'_ /|\JAV'Za7e


Both components are printed to the output and a newline is automatically placed between the components

Try it at MATL Online

# Python 2, 128 127 bytes

-1 byte thanks to Rod (use multiplication of tuple ('_'*10,) to avoid a declaration).

print''.join('0'<c<':'and' '*int(c)or c for c in'''2o
7o
4o
%s
/ 9\__
|3J6|2\\
|4A5|2|
|5V4|2|
|6A3|__/
\%s/'''%(('_'*10,)*2))


Try it online!

Note: that double backslash is needed before the line feed.

Everything between the ''' and ''' is a single string, the two %s are formatters which get replaced by the content of the trailing %(...) tuple, which in turn contains two copies of '_'*10 via the tuple multiplication (...)*2. The '_'*10 performs string multiplication to yield '__________'.

The code traverses the characters, c, of that whole string using for c in '''... and creates a new string by joining (join(...))
either the number of spaces identified by c, int(c), if c is a digit
or c itself
- being a digit is identified by '0'<c<':' to save over c.isdigit().

• You can replace u,u with ('_'*10,)*2 and drop the u declaration
– Rod
Feb 24, 2017 at 16:23
• Oh, nice I did look at that and think there was a way - thanks @Rod! Feb 24, 2017 at 16:53

# Java 8, 294289 248 bytes

Golfed:

()->{String s="";for(char c:"\u026F\n\u076F\n\u046F\n __________\n/\u0A5C__\n|\u034A\u067C\u025C\n|\u0441\u057C\u027C\n|\u0556\u047C\u027C\n|\u0641\u037C__/\n\\__________/".toCharArray()){for(int i=0;i<c>>8;++i)s+=' ';s+=(char)(c&255);}return s;}


In the spirit of , this does not hard-code the string to output. Instead, it makes use of the fact that there are many cases of multiple spaces followed by a printable character. It encodes the number of spaces that precede a character in the high-order byte of the character, with the actual ASCII character in the low-order byte.

Ungolfed:

import java.util.function.*;

public class DrinkYourMorningCoffee {

public static void main(String[] args) {
System.out.println(f(
() -> {
String s = "";
for (char c : "\u026F\n\u076F\n\u046F\n __________\n/\u0A5C__\n|\u034A\u067C\u025C\n|\u0441\u057C\u027C\n|\u0556\u047C\u027C\n|\u0641\u037C__/\n\\__________/".toCharArray()) {
for (int i = 0; i < c >> 8; ++i) {
s += ' ';
}
s += (char) (c & 255);
}
return s;
}
));
}

private static String f(Supplier<String> s) {
return s.get();
}
}

• I think it would be better to encode the number of leading spaces before a character in the high byte. So, an 'A' preceded by six spaces would be encoded as \u0641. Feb 26, 2017 at 18:03
• @DavidConrad why not do both? There are no more than ten consecutive repetitions anywhere in the string, and decimal ten fits in hex F. It should be possible to fit both in there.
– user18932
Feb 26, 2017 at 18:06
• That's a good point. Feb 26, 2017 at 18:10
• @DavidConrad it ended up making the file size bigger due to the second loop I had to add. But I did save a few bytes by converting hex to decimal constants. Losing the 0x prefix helped.
– user18932
Feb 26, 2017 at 18:57
• There are two encodings that look useful: number of spaces, and number of repetitions. You were correct: encoding the number of spaces is a net gain. I was also able to golf some of the other code (if is redundant, for example) and shave off around 1/6th of the size.
– user18932
Feb 26, 2017 at 19:16

\d
$*  (there's a trailing space at the end) Try it online! The principle is still having a "compressed" string from which the cup of coffee can be reconstructed by substitutions. Trying different substitutions it turned out that the only ones worth doing are: • = turns into __________ (10 underscores) • any digit turns into that number of spaces # Common Lisp, 125123122120 114 bytes (format t"~3@{~vto ~} ~10@{_~} /~11t\\__ | J~11t| \\ ~2@{|~5t~a~11t| | ~}|~7tA |__/ \\~10{_~}/"2 7 4'A" V"1)  I saved 6 bytes, using idea of just putting enters in string instead of ~&s. Ideas for improvement welcomed. # Python3, 206 bytes print(' o\n'+7*' '+'o\n'+4*' '+'o\n'+' '+10*'_'+'\n'+'/'+10*' '+'\__\n'+'|'+3*' '+'J'+6*' '+'| \\\n'+'|'+4*' '+'A'+5*' '+'| |\n'+'|'+5*' '+'V'+4*' '+'| |\n'+'|'+6*' '+'A'+3*' '+'|__/\n'+'\\'+10*'_'+'/')  • So many space characters… Better declare a s=' ' variable and use it. Feb 24, 2017 at 14:36 • Also wouldn't hurt to predefine newline Feb 24, 2017 at 14:37 • Hardcoding the output is shorter Feb 24, 2017 at 14:41 • @WheatWizard, I don't think so. There is only a single solitary newline. The others are inside strings, so using a variable would also need a concatenation operator. And 'o\n' has the same length as 'o'+n. Feb 24, 2017 at 15:50 • @manatwork One could: print(*(' o',7*' '+'o',4*' '+'o',' '+10*'_','/'+10*' '+'\__','|'+3*' '+'J'+6*' '+'| \\','|'+4*' '+'A'+5*' '+'| |','|'+5*' '+'V'+4*' '+'| |','|'+6*' '+'A'+3*' '+'|__/','\\'+10*'_'+'/'),sep='\n') or for x in(' o',7*' '+'o',4*' '+'o',' '+10*'_','/'+10*' '+'\__','|'+3*' '+'J'+6*' '+'| \\','|'+4*' '+'A'+5*' '+'| |','|'+5*' '+'V'+4*' '+'| |','|'+6*' '+'A'+3*' '+'|__/','\\'+10*'_'+'/'):print(x), both are 197. Still longer than a 136 hard code. Feb 24, 2017 at 16:02 # Pyth, 80 bytes r" o 7 o 4 o 10_ /10 \__ |3 J6 | \\ |4 A5 | | |5 V4 | | |6 A3 |__/ \\10_/"9  Online interpreter available here. Simple run-length decoding. # C - 179 Solution with extensive use of format string: void f(){printf("%1$3c\n%1$8c\n%1$5c\n%2$11s\n/%3$13s\n|%4$4c%5$7c%6$3c\n|%7$5c%5$6c%5$3c\n|%8$6c%5$5c%5$3c\n|%7$7c%5$4c__/\n\\%2$s/\n",'o',"__________","\\__",74,'|',92,65,86);}


Here is a more readable version:

void f() {
printf("%1$3c\n" "%1$8c\n"
"%1$5c\n" "%2$11s\n"
"/%3$13s\n" "|%4$4c%5$7c%6$3c\n"
"|%7$5c%5$6c%5$3c\n" "|%8$6c%5$5c%5$3c\n"
"|%7$7c%5$4c__/\n"
6o
3o
9
/44\\__
|2J5|1\\
|3A4|1|
|4V3|1|
|5A2|__/
\\9/"[$i++];)echo$c>0?str_repeat(" _"[$c>8],$c+1):$c;  This looks a lot like Arnauld´s answer - and does pretty much the same. Run with -r. # zsh, 86 bytes printf "^_<8b>^H^@^@^@^@^@^B^CSPÈçR^@^A^P^CJÆÃ^A<97>¾^B^\Ä^@¹5@Ú^KÂ^E2cÀ|^EG^X¿^FÂW^HCæÃTÔÄÇësÅÀ^L^Fq^@<92>}ý^?{^@^@^@"|zcat  Explanation: that string is the gzip-compressed java cup ascii art. I use printf, because with echo, zcat prints a warning, and echo -e is one character longer. It doesn't work with bash or sh, because they think it's a binary file. Since you can't effectively paste that output from the browser, here's a usable file. # Java 9 / JShell, 299 bytes ()->{String s="";BigInteger b=new BigInteger("43ljxwxunmd9l9jcb3w0rylqzbs62sy1zk7gak5836c2lv5t36ej6682n2pyucm7gkm9bkfbn4ttn0gltbscvbttifvtdfetxorj6mmy3mt6r3",36);while(!b.equals(BigInteger.ZERO)){int x=b.intValue()&0x3ff;for(int i=0;i<x>>7;i++)s+=' ';s+=(char)(x&0x7f);b=b.shiftRight(10);}return s;}  Ungolfed: () -> { String s = ""; BigInteger b = new BigInteger( "43ljxwxunmd9l9jcb3w0rylqzbs62sy1zk7gak5836c2lv5t36ej6682n2pyucm7gkm9bkfbn4ttn0gltbscvbttifvtdfetxorj6mmy3mt6r3", 36); while (!b.equals(BigInteger.ZERO)) { int x = b.intValue() & 0x3ff; for (int i = 0; i < x >> 7; i++) s+=' '; s += (char)(x&0x7f); b = b.shiftRight(10); } return s; }  Usage in JShell: Supplier<String> golf = <lambda expression> System.out.println(golf.get())  Encodes each character as ten bits consisting of a count of the number of spaces before the character in the high three bits following by the code point in the low seven bits. (Since there are only three bits for the count it can't represent more than seven consecutive spaces, and there are ten spaces at one point in the string. These are encoded as a count of six, followed by a space, and then a count of three followed by the next character.) Sadly, it loses to this trivial 140-byte Java solution: ()->" o\n o\n o\n __________\n/ \\__\n| J | \\\n| A | |\n| V | |\n| A |__/\n\\__________/"  # 05AB1E, 85 bytes •1d'uì[_ÍpH»Ð]jŠ$ÿ{É˜ß|ªpå±W¾Ö:ÞjÇ&@è\$´Öàˆå]Á¢šBg¦ï&-ã¥ønØ7Ñà'?•9B8ÝJ"o _/\|JAV"‡15ô»


Try it online!

# 05AB1E, 67 65 bytes

'oƵ®Sú»•BÇÉι!т³ÂSˆtTóV”PHAëdÀāåΘý”å.@=Ø₂)øUqrÖ½ñ*•“_ |
\/AVJ“ÅвJ»


Pretty straight-forward approach which most likely has room for improvement.

Try it online.

Explanation:

'o            '# Push an "o"
Ƶ®           # Push compressed integer 274
S          # Convert it to a list of digits: [2,7,4]
ú         # Prepend that many leading spaces to the "o": ["  o","       o","   o"]
»        # Join them by newlines: "  o\n       o\n   o"
•BÇÉι!т³ÂSˆtTóV”PHAëdÀāåΘý”å.@=Ø₂)øUqrÖ½ñ*•
# Push compressed integer 21514733101393053622421734761651860518184887728386739717403046550918945532453802933506348033205129
“_ |\n\/AVJ“  # Push string "_ |\n\/AVJ"
Åв           # Convert the integer to base-"_ |\n\/AVJ"
# (builtin for converting to base string-length, and index those into the string)
J          # Join the characters to a single string
»         # And join the two strings on the stack by a newline
# (after which the result is output implicitly as result)


See this 05AB1E tip of mine (section How to compress large integers?) to understand why Ƶ® is 274 and •BÇÉι!т³ÂSˆtTóV”PHAëdÀāåΘý”å.@=Ø₂)øUqrÖ½ñ*• is 21514733101393053622421734761651860518184887728386739717403046550918945532453802933506348033205129.

# Vyxal, 108 bytes

274ƛI\o꘍,;_ð\_₀*+,\/₀꘍\__+,\|3꘍\J6꘍\|ðd\\Wṅ,\|4꘍\A5꘍| m++,\|5꘍\V4꘍| m++,\|6꘍\A3꘍|__/++,\\\_₀*\/++,


Try it Online!

# 05AB1E, 65 bytes

'oƵ®Sú»•BÇÉι₆₅8á·ŸˆE©/LнDË9Ï0³†₃þC¤ā¸·Í¨z©´Kˆι₃BÄ•"_
/\|JAV"ÅвJ»


Try it online!

# Pure Åв compression (compressing the bubbles too), 67 bytes

•5m™ÔćtвÃÔηŒćõ»θpøàÄqŠZ₂rÓ¢γβÞĆ?úËÄÌUöÉн%Cñí₄‡±ÁkÆùi•"o
_/\|JAV"Åв


Try it online! Outputs as a list of characters.

# Λ canvas, 109 bytes

'oƵ®Sú»6…/||º4Λ12„\_5×«º2Λ2„/|0Λ4„|_º2Λ4„/|º0Λ4„\ º6Λ4'|4Λ5'|0Λ4ð2Λ4„ _º6Λ2'\0Λ2ð6ΛT'_6Λ3…_  4Λ4ð2Λ4'îáu3.Λ»


Try it online!

# Canvas, 59 bytes

？Eα)⁶^↑^jo↕ｊα◂9Ｓ⁷YNＵ>AｇＸ↷ＵＥ╬ｑ\ｄ7ｆ²ｔ}ＭøＱ＃║Ｍ<Y5２ｄｐc３Ｉ‰０Ｃ5Ｇ９－‟
`

Try it online!