Outputting ordinal numbers (1st, 2nd, 3rd)

Question

I would like to generate (as a return result of a function, or simply as the output of a program) the ordinal suffix of a positive integer concatenated to the number.

Samples:

1st  
2nd  
3rd  
4th  
...  
11th  
12th  
13th  
...  
20th  
21st
22nd
23rd
24th

And so on, with the suffix repeating the initial 1-10 subpattern every 10 until 100, where the pattern ultimately starts over.

The input would be the number and the output the ordinal string as shown above.

What is the smallest algorithm for this?

Answer 1

Perl, 37 + 1 characters

s/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg

This is a regexp substitution that appends the appropriate ordinal suffix to any numbers in $_ that are not already followed by a letter. To apply it to file input, use the p command line switch, like this:

perl -pe 's/1?\d\b/$&.((0,st,nd,rd)[$&]||th)/eg'

This is a complete Perl program that reads input from stdin and writes the processed output to stdout. The actual code is 37 chars long, but the p switch counts as one extra character.

Sample input:

This is the 1 line of the sample input...
...and this is the 2.
1 2 3 4 5 6 7 8 9 10
11 12 13 14 15 16 17 18 19 20
21 22 23 24 25 26 27 28 29 30
101 102 103 104 105 106 107 108 109 110

Output:

This is the 1st line of the sample input...
...and this is the 2nd.
1st 2nd 3rd 4th 5th 6th 7th 8th 9th 10th
11th 12th 13th 14th 15th 16th 17th 18th 19th 20th
21st 22nd 23rd 24th 25th 26th 27th 28th 29th 30th
101st 102nd 103rd 104th 105th 106th 107th 108th 109th 110th

Numbers already followed by letters will be ignored, so feeding the output again through the filter won't change it. Spaces, commas and periods between numbers are not treated specially, so they're assumed to separate numbers like any other punctuation. Thus, e.g. 3.14159 becomes 3rd.14159th.

How does it work?

• First, this is a global regexp replacement (s///g). The regexp being matched is 1?\d\b, where \d matches any digit and \b is a zero-width assertion matching the boundary between an alphanumeric and a non-alphanumeric character. Thus, 1?\d\b matches the last digit of any number, plus the previous digit if it happens to be 1.

• In the substitution, which is evaluated as Perl code due to the /e switch, we take the matched string segment ($&) and append (.) to it the suffix obtained by using $& itself as an integer index to the list (0,st,nd,rd); if this suffix is zero or undefined (i.e. when $& is zero or greater than three), the || operator replaces it with th.

---

Edit: If the input is restricted to a single integer, then this 35 character solution will suffice:

s/1?\d$/$&.((0,st,nd,rd)[$&]||th)/e

Answer 2

Python 2, 49 bytes

lambda n:`n`+'tsnrhtdd'[n%5*(n%100^15>4>n%10)::4]

An anonymous function. A full program would be counted at 55 bytes.

'tsnrhtdd'[i::4] encodes the suffixes th st nd rd for values of i from 0 to 3. Given this, all we need is a way to map the values of n to the index of the corresponding suffix, i. A straightforward expression that works is (n%10)*(n%10<4 and 10<n%100<14). We can easily shorten this by dropping the first set of parentheses and observing that n%5 gives the same results as n%10 for the values of n with the special suffixes. With a bit of trial and error, one may also shorten 10<n%100<14 to n%100^15>4, which can be chained with the other conditional to save even more bytes.

Answer 3

Mathematica 39 45 bytes

Note: In recent versions of Mathematica, asking for the nth part of p, where p is undefined, generates an error message, but returns the correct answer anyway. I've added Quiet to prevent the error message from printing.

Quiet@StringSplit[SpokenString[p[[#]]]][[2]]&

Usage

Quiet@StringSplit[SpokenString[p[[#]]]][[2]] &[31]

31st

Quiet@StringSplit[SpokenString[p[[#]]]][[2]] &/@Range[21]

{"1st", "2nd", "3rd", "4th", "5th", "6th", "7th", "8th", "9th", "10th", "11th", "12th", "13th", "14th", "15th", "16th", "17th", "18th", "19th", "20th", "21st"}

---

How it works

SpokenString writes out an any valid Mathematica expression as it might be spoken. Below are two examples from the documentation for SpokenString,

SpokenString[Sqrt[x/(y + z)]]

"square root of the quantity x over the quantity y plus z" *)

SpokenString[Graphics3D[Sphere[{{0, 0, 0}, {1, 1, 1}}]], "DetailedGraphics" -> True]

"a three-dimensional graphic consisting of unit spheres centered at 0, 0, 0 and 1, 1, 1"

---

Now, for the example at hand,

Quiet@SpokenString[p[[#]]] &[31]

"the 31st element of p"

Let's represent the above string as a list of words:

StringSplit[%]

{"the", "31st", "element", "of", "p"}

and take the second element...

%[[2]]

31st

Answer 4

Python, 68 characters

i=input()
k=i%10
print"%d%s"%(i,"tsnrhtdd"[(i/10%10!=1)*(k<4)*k::4])

Answer 5

Ruby, 60

It's not as good as the Perl entry, but I figured I'd work on my Ruby skills.

def o(n)n.to_s+%w{th st nd rd}[n/10%10==1||n%10>3?0:n%10]end

Function takes one integer argument, n, and returns a string as the ordinal form.

Works according to the following logic:
If the tens digit is a 1 or the ones digit is greater than 3 use the suffix 'th'; otherwise find the suffix from the array ['th', 'st', 'nd', 'rd'] using the final digit as the index.

Answer 6

Javascript (ES6) 50 44 Bytes (non-competing)

a=>a+=[,"st","nd","rd"][a.match`1?.$`]||"th"

Notes

• takes imput as a string
• Removed 6 bytes, thanks @user81655

Answer 7

Javascript, 68 71

function o(n){return n+([,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th')}

Joint effort with ItsCosmo.

EDIT: Wasn't working properly with numbers > 100

Answer 8

Haskell, 95 chars

h=foldr g"th".show
g '1'"th"="1st"
g '2'"th"="2nd"
g '3'"th"="3rd"
g '1'[x,_,_]='1':x:"th"
g a s=a:s

Testing:

*Main> map h [1..40]
["1st","2nd","3rd","4th","5th","6th","7th","8th","9th","10th","11th","12th","13t
h","14th","15th","16th","17th","18th","19th","20th","21st","22nd","23rd","24th",
"25th","26th","27th","28th","29th","30th","31st","32nd","33rd","34th","35th","36
th","37th","38th","39th","40th"]

Must be loaded with -XNoMonomorphismRestriction.

Answer 9

JavaScript, 64 characters (ES3) or 47 characters (ES6)

ES3 (64 characters):

function(n){return n+=[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'}

ES6 (47 characters):

n=>n+=[,'st','nd','rd'][n%100>>3^1&&n%10]||'th'

Explanation

The expression n % 100 >> 3 ^ 1 evaluates to 0 for any positive n ending with digits 08–15. Thus, for any n mod 100 ending in 11, 12, or 13, the array lookup returns undefined, leading to a suffix of th.

For any positive n ending in other digits than 08–15, the expression n % 100 >> 3 ^ 1 evaluates to a positive integer, invoking the expression n % 10 for array lookup, returning st,nd, or rd for n which ends with 1, 2, or 3. Otherwise, th.

Answer 10

Golfscript, 34 characters

~.10/10%1=!1$10%*.4<*'thstndrd'2/=

Answer 11

APL (Dyalog Unicode), 38 36 bytes

Thanks to ngn for fixing a bug while maintaining byte count.

Anonymous tacit prefix function. Requires ⎕IO (Index Origin) set to 0, which is default on many systems. Even works for 0!

⍕,{2↑'thstndrd'↓⍨2×⊃⍵⌽∊1 0 8\⊂10↑⍳4}

Try it online!

{…} anonymous lambda; ⍵ is argument:

 ⍳4 first four ɩndices; [0,1,2,3]

 10↑ take first ten elements from that, padding with zeros: [0,1,2,3,0,0,0,0,0,0]

 ⊂ enclose to treat as single element; [[0,1,2,3,0,0,0,0,0,0]]

 1 0 8\ expand to one copy, a prototypical copy (all-zero), eight copies;
  [[0,1,2,3,0,0,0,0,0,0],
   [0,0,0,0,0,0,0,0,0,0],
   [0,1,2,3,0,0,0,0,0,0],
   [0,1,2,3,0,0,0,0,0,0],
   ⋮ (5 more)
   [0,1,2,3,0,0,0,0,0,0]]

 ∊ ϵnlist (flatten);
  [0,1,2,3,0,0,0,0,0,0,
   0,0,0,0,0,0,0,0,0,0,
   0,1,2,3,0,0,0,0,0,0,
   0,1,2,3,0,0,0,0,0,0,
   ⋮ (50 more)
   0,1,2,3,0,0,0,0,0,0]

 ⍵⌽ cyclically rotate left as many steps as indicated by the argument

 ⊃ pick the first number (i.e. the argument-mod-100'th number)

 2× multiply two by that (gives 0, 2, 4, or 6)

 'thstndrd'↓⍨drop that many characters from this string

 2↑ take the first two of the remaining characters

⍕, concatenate the stringified argument to that

Answer 12

C#, 62 bytes

n=>n+(n/10%10==1||(n%=10)<1||n>3?"th":n<2?"st":n<3?"nd":"rd");

Full program and verification:

using System;

namespace OutputOrdinalNumbers
{
    class Program
    {
        static void Main(string[] args)
        {
            Func<int,string>f= n=>n+(n/10%10==1||(n%=10)<1||n>3?"th":n<2?"st":n<3?"nd":"rd");

            for (int i=1; i<=124; i++)
                Console.WriteLine(f(i));
        }
    }
}

Answer 13

J - 44 char

Nothing in J? This is an outrage!

(":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:])

Explained (note that 1 is boolean true in J and 0 is false):

• 10 10(...)/@#:] - First we take the argument (]) and find the tens and ones digit (10 10 #:). Then, we will insert (...) between the two.
• (]*[(~:*])4>]) - In this subexpression, but not the innermost one, ] will point to the ones digit and [ the tens digit.
• [(~:*])4>] - ~: is J for "not-equals", so this takes the result of 4>] (i.e. whether or not the one digit is less than 4) and multiplies it by the result of tens ~: (4>]). Why would anyone do this? Consider the following:
  • If we are checking 10, 11, 12, 13, then tens is 1 (we are in the teens) and ones is less than 4, so tens ~: (4>]) is false and the result is 0*1 = 0.
  • If we are any other one of {X0, X1, X2, X3}, then clearly tens ~: (4>]) is true and we get out 1*1 = 1.
  • If ones is greater than four, then 4>] was 0 and it doesn't matter what happens to the test anymore, we will get 0 out regardless.
  • So to summarize, [(~:*])4>] is 1 if we are in {X0, X1, X2, X3} but not in the teens, and 0 otherwise.
• ]* - Finally we multiply that result by the ones digit. So this product will be 0 if the number deserves a 'th' suffix, else its value.
• th`st`nd`rd{::~ - We use the modified ones-digit from above to index the list of suffixes. 0 gets 'th', 1 gets 'st', and so on.
• ":, - Finally, take the original number, convert it to a string (":), and then prepend it to the suffix.

Usage is obvious, though as-is the verb can only take one ordinal, not a list.

   (":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:]) 112         NB. single use
112th
   (":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:]) 1 2 3 4 5   NB. doing it wrong
|length error
|       (":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:])1 2 3 4 5
   NB. i.5 10   makes a 5x10 grid of increasing integers
   NB. &.>      to operate on each integer separately, and box the result after
   (":,th`st`nd`rd{::~10 10(]*[(~:*])4>])/@#:])&.> i.5 10   NB. all better
+----+----+----+----+----+----+----+----+----+----+
|0th |1st |2nd |3rd |4th |5th |6th |7th |8th |9th |
+----+----+----+----+----+----+----+----+----+----+
|10th|11th|12th|13th|14th|15th|16th|17th|18th|19th|
+----+----+----+----+----+----+----+----+----+----+
|20th|21st|22nd|23rd|24th|25th|26th|27th|28th|29th|
+----+----+----+----+----+----+----+----+----+----+
|30th|31st|32nd|33rd|34th|35th|36th|37th|38th|39th|
+----+----+----+----+----+----+----+----+----+----+
|40th|41st|42nd|43rd|44th|45th|46th|47th|48th|49th|
+----+----+----+----+----+----+----+----+----+----+

Answer 14

PowerShell, 92

process{"$_$(switch -r($_){"(?<!1)1$"{'st'}"(?<!1)2$"{'nd'}"(?<!1)3$"{'rd'}default{'th'}})"}

Works with one number per line of input. Input is given through the pipeline. Making it work for only a single number doesn't reduce the size.

Answer 15

Mathematica 29 + 5 = 34 bytes

SpokenStringDump`SpeakOrdinal

+5 bytes because Speak function must be called before using this built-in.

Usage

SpokenStringDump`SpeakOrdinal[1]

"1st "

SpokenStringDump`SpeakOrdinal[4707]

"4,707th "

Answer 16

JavaScript (Node.js), 51 bytes

credit to @KevinCruijssen for improving the answer

n=>n+=[,'st','nd','rd'][~~(n/10%10)-1?n%10:0]||'th'

Try it online!

---

Explanation :

n =>                      // input
    n+=[,'st','nd','rd']     // the array of suffices add everything to n
        [~~(n/10%10)-1?n%10:0] // the full magic 
    || 'th'               // works for everything except 1 , 2 , 3 just like what you want

---

Answer 17

R, 79 76 bytes

Since there is no R solution yet... no tricks here, basic vector indexing, golfed down 3 chars thanks to Giuseppe. Previously tried index: [1+(x%%10)-(x%%100==11)] and [1+(x%%10)*(x%%100!=11)].

function(x)paste0(x,c("th","st","nd","rd",rep("th",6))[1+x%%10*!x%%100==11])

Try it online!

With substr, 79 bytes:

function(x,y=1+2*min(x%%10-(x%%100==11),4))paste0(x,substr("thstndrdth",y,y+1))

Try it online!

Answer 18

K - 44 char

It so happens that this is exactly as long as the J, and works in almost the same way.

{x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:}

Explained:

• x$: - First, we convert the operand x into a string, and then assign that back to x. We will need its string rep again later, so doing it now saves characters.
• .:' - Convert (.:) each (') digit back into a number.
• -2#0, - Append a 0 to the front of the list of digits (in case of single-digit numbers), and then take the last two.
• {y*(y<4)*~1=x}. - Use the two digits as arguments x and y to this inner function, which returns y if y is less than 4 and x is not equal to 1, otherwise 0.
• `th`st`nd`rd@ - Index the list of suffixes by this result.
• x,$ - Convert the suffix from symbol to string, and append it to the original number.

Usage:

  {x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:} 3021
"3021st"
  {x,$`th`st`nd`rd@{y*(y<4)*~1=x}.-2#0,.:'x$:}' 1 2 3 4 11 12 13 14  /for each in list
("1st"
 "2nd"
 "3rd"
 "4th"
 "11th"
 "12th"
 "13th"
 "14th")

Answer 19

OCaml

I'm pretty new to OCaml, but this is the shortest i could get.

let n x =
   let v = x mod 100 and string_v = string_of_int x in
   let z = v mod 10 in
   if v=11 || v=12 || v=13 then string_v^"th" 
   else if v = 1 || z = 1 then string_v^"st" else if v = 2 || z = 2 then string_v^"nd" else if v = 3 || z = 3 then string_v^"rd" else string_v^"th";;

I created a function n that takes a number as a parameter and does the work. Its long but thought it'd be great to have a functional example.

Answer 20

C - 95 83 characters

main(n,k){scanf("%d",&n);k=(n+9)%10;printf("%d%s\n",n,k<3?"st\0nd\0rd"+3*k:"th");}

Degolfed:

main(n,k)
{
    scanf("%d",&n);
    k=(n+9)%10; // xx1->0, xx2->1, xx3->2
    printf("%d%s\n",n,k<3?"st\0nd\0rd"+3*k:"th");
}

We could do k=(n-1)%10 instead of adding 9, but for n=0 we would get an incorrect behaviour, because in C (-1)%10 evaluates to -1, not 9.

Answer 21

Javascript, 75

function s(i){return i+'thstndrd'.substr(~~(i/10%10)-1&&i%10<4?i%10*2:0,2)}

Answer 22

PHP, 151

I know that this program is not comparable to the others. Just felt like giving a solution.

<?$s=explode(' ',trim(fgets(STDIN)));foreach($s as$n){echo$n;echo(int)(($n%100)/10)==1?'th':($n%10==1?'st':($n%10==2?'nd':($n%10==3?'rd':'th')))."\n";}

Answer 23

Scala 86

def x(n:Int)=n+{if(n%100/10==1)"th"else(("thstndrd"+"th"*6).sliding(2,2).toSeq(n%10))}

Scala 102:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".take(n+1)%5*2.drop(n%5*2))else n+"th"

102 as well:

def x(n:Int)=if((n%100)/10==1)n+"th"else if(n%10<4)n+("thstndrd".sliding(2,2).toSeq(n%10))else n+"th"

ungolfed:

def x (n: Int) =
  n + { if (((n % 100) / 10) == 1) "th" 
        else (("thstndrd"  + ("th"  * 6)).sliding (2, 2).toSeq (n % 10))
      }

Answer 24

PHP, 98 bytes

function c($n){$c=$n%100;$s=['','st','nd','rd'];echo$c>9&&$c<14||!$s[$c%10]?$n.'th':$n.$s[$c%10];}

The 11-13 bit is killing me here. Works for any integer $n >= 0.

For any integer $n:

PHP, 103 bytes

function c($n){$c=abs($n%100);$s=['','st','nd','rd'];echo$c>9&&$c<14||!$s[$c%10]?$n.'th':$n.$s[$c%10];}

Answer 25

Python, 88 84 bytes

lambda x:x+((('th','st','nd','rd')+('th',)*6)[int(x[-1])]if('0'+x)[-2]!='1'else'th')

Ungolfed:

def y(x):
    return x + (('th', 'st', 'nd', 'rd') + ('th', ) * 6)[int(x[-1])] if ('0' + x)[-2] != '1' else 'th')

lambda x defines an anonymous function with parameter x. ((('th','st','nd','rd')+('th',)*6)[int(x[-1])] defines a tuple of the endings for numbers less than 10, the 0-th element is for 0, and so forth. the if ('0'+x)[-2] != '1' checks if there is 11, 12, or a 13 to fix, and adds then else 'th' adds th instead of st, rd, or nd.

Answer 26

Common Lisp, 63

(format t"~a~a"i(let((x(format()"~:r"i)))(subseq x(-(length x)2))))

Input is i. There's probably a more clever way to golf this.

Answer 27

C: 95 characters

A ridiculously long solution:

n;main(){scanf("%d",&n);printf("%d%s",n,n/10%10-1&&(n=n%10)<4&&n?n>2?"rd":n<2?"st":"nd":"th");}

It needs to be mangled more.

Answer 28

Javascript, 75

function o(s){return s+((0|s/10%10)==1?"th":[,"st","nd","rd"][s%10]||"th")}

Answer 29

Pyth, 33 bytes

+Q@.>+c"stndrd"2*]"th"7 1?qh`Q\1Z

Try it online!

Answer 30

Procedural Footnote Language, 39 bytes

[1]
[PFL1.0]
[1] [ORD:[INPUT]]
[PFLEND]

The BODY section of the document contains a reference to footnote [1]; footnote [1] contains the [ORD]inal representation of the [INPUT].

Answer 31

Julia 0.6, 68 bytes

f(x,m=x%10,p=x%100-m)="$x$(["th","st","nd","rd"][m>3||p==10?1:m+1])"

Try it online!

Straightforward remainder rules in Julia.

Answer 32

Python 2.7, 137 chars

def b(n):
 for e,o in[[i,'th']for i in['11','12','13']+list('4567890')]+[['2','nd'],['1','st'],['3','rd']]:
  if n.endswith(e):return n+o

n should be a string

I know I'm beaten by the competition here already, but I thought I'd provide my idea anyways

this just basically generates a list of key, value pairs with number (as a string) ending e and the ordinal o. It attempts to match 'th' first (hence why I didn't use a dictionary), so that it won't accidentally return 'st', for example, when it should be 'th'. This will work for any positive integer

Answer 33

Haxe, 83 chars

function o(n:Int){return ['th','st','nd','rd'][n%100>20||n%100<4?n%10>3?0:n%10:0];}

Based on the javascript version, fixed for n > 100 as well.

Answer 34

Oracle SQL 11.2, 101 bytes

SELECT:1||DECODE(ROUND(MOD(:1,100),-1),10,'th',DECODE(MOD(:1,10),1,'st',2,'nd',3,'rd','th'))FROM DUAL

Answer 35

Javascript ES6, 52 chars

n=>n+(!/1.$/.test(--n)&&'snr'[n%=10]+'tdd'[n]||'th')

Answer 36

Scala 2.11.7, 151 Chars

def o(x:Int*)=x map{
case a if(a%10)==1&&a%100!=11=>a+"st"
case b if(b%10)==2&&b%100!=12=>b+"nd"
case c if(c%10)==3&&c%100!=13=>c+"rd"
case e=>e+"th"
}

Usage

   o(1,2,3,4) -> ArrayBuffer("1st", "2nd", "3rd", "4th")

Answer 37

Java 7, 91 81 bytes

String d(int n){return n+(n/10%10==1|(n%=10)<1|n>3?"th":n<2?"st":n<3?"nd":"rd");}

Port from @adrianmp's C# answer.

Old answer (91 bytes):

String c(int n){return n+((n%=100)>10&n<14?"th":(n%=10)==1?"st":n==2?"nd":n==3?"rd":"th");}

Ungolfed & test code:

Try it here.

class M{
  static String c(int n){
    return n + ((n%=100) > 10 & n < 14
                ?"th"
                : (n%=10) == 1
                   ? "st"
                   : n == 2
                      ? "nd"
                      : n == 3
                         ? "rd"
                         :"th");
  }

  static String d(int n){
    return n + (n/10%10 == 1 | (n%=10) < 1 | n > 3
                 ? "th"
                 : n < 2
                    ? "st"
                    : n < 3
                       ? "nd"
                       :"rd");
  }

  public static void main(String[] a){
    for(int i = 1; i < 201; i++){
      System.out.print(c(i) + ", ");
    }
    System.out.println();
    for(int i = 1; i < 201; i++){
      System.out.print(d(i) + ", ");
    }
  }
}

Output:

1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th, 20th, 21st, 22nd, 23rd, 24th, 25th, 26th, 27th, 28th, 29th, 30th, 31st, 32nd, 33rd, 34th, 35th, 36th, 37th, 38th, 39th, 40th, 41st, 42nd, 43rd, 44th, 45th, 46th, 47th, 48th, 49th, 50th, 51st, 52nd, 53rd, 54th, 55th, 56th, 57th, 58th, 59th, 60th, 61st, 62nd, 63rd, 64th, 65th, 66th, 67th, 68th, 69th, 70th, 71st, 72nd, 73rd, 74th, 75th, 76th, 77th, 78th, 79th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, 90th, 91st, 92nd, 93rd, 94th, 95th, 96th, 97th, 98th, 99th, 100th, 101st, 102nd, 103rd, 104th, 105th, 106th, 107th, 108th, 109th, 110th, 111th, 112th, 113th, 114th, 115th, 116th, 117th, 118th, 119th, 120th, 121st, 122nd, 123rd, 124th, 125th, 126th, 127th, 128th, 129th, 130th, 131st, 132nd, 133rd, 134th, 135th, 136th, 137th, 138th, 139th, 140th, 141st, 142nd, 143rd, 144th, 145th, 146th, 147th, 148th, 149th, 150th, 151st, 152nd, 153rd, 154th, 155th, 156th, 157th, 158th, 159th, 160th, 161st, 162nd, 163rd, 164th, 165th, 166th, 167th, 168th, 169th, 170th, 171st, 172nd, 173rd, 174th, 175th, 176th, 177th, 178th, 179th, 180th, 181st, 182nd, 183rd, 184th, 185th, 186th, 187th, 188th, 189th, 190th, 191st, 192nd, 193rd, 194th, 195th, 196th, 197th, 198th, 199th, 200th, 
1st, 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th, 10th, 11th, 12th, 13th, 14th, 15th, 16th, 17th, 18th, 19th, 20th, 21st, 22nd, 23rd, 24th, 25th, 26th, 27th, 28th, 29th, 30th, 31st, 32nd, 33rd, 34th, 35th, 36th, 37th, 38th, 39th, 40th, 41st, 42nd, 43rd, 44th, 45th, 46th, 47th, 48th, 49th, 50th, 51st, 52nd, 53rd, 54th, 55th, 56th, 57th, 58th, 59th, 60th, 61st, 62nd, 63rd, 64th, 65th, 66th, 67th, 68th, 69th, 70th, 71st, 72nd, 73rd, 74th, 75th, 76th, 77th, 78th, 79th, 80th, 81st, 82nd, 83rd, 84th, 85th, 86th, 87th, 88th, 89th, 90th, 91st, 92nd, 93rd, 94th, 95th, 96th, 97th, 98th, 99th, 100th, 101st, 102nd, 103rd, 104th, 105th, 106th, 107th, 108th, 109th, 110th, 111th, 112th, 113th, 114th, 115th, 116th, 117th, 118th, 119th, 120th, 121st, 122nd, 123rd, 124th, 125th, 126th, 127th, 128th, 129th, 130th, 131st, 132nd, 133rd, 134th, 135th, 136th, 137th, 138th, 139th, 140th, 141st, 142nd, 143rd, 144th, 145th, 146th, 147th, 148th, 149th, 150th, 151st, 152nd, 153rd, 154th, 155th, 156th, 157th, 158th, 159th, 160th, 161st, 162nd, 163rd, 164th, 165th, 166th, 167th, 168th, 169th, 170th, 171st, 172nd, 173rd, 174th, 175th, 176th, 177th, 178th, 179th, 180th, 181st, 182nd, 183rd, 184th, 185th, 186th, 187th, 188th, 189th, 190th, 191st, 192nd, 193rd, 194th, 195th, 196th, 197th, 198th, 199th, 200th,