Problem:
Score : 300 points

Problem Statement

1000000000000001 dogs suddenly appeared under the roof of Roger's house, all of which he decided to keep. The dogs had been numbered 1 through 1000000000000001, but he gave them new names, as follows:

the dogs numbered 1,2,\cdots,26 were respectively given the names a, b, ..., z;

the dogs numbered 27,28,29,\cdots,701,702 were respectively given the names aa, ab, ac, ..., zy, zz;

the dogs numbered 703,704,705,\cdots,18277,18278 were respectively given the names aaa, aab, aac, ..., zzy, zzz;

the dogs numbered 18279,18280,18281,\cdots,475253,475254 were respectively given the names aaaa, aaab, aaac, ..., zzzy, zzzz;

the dogs numbered 475255,475256,\cdots were respectively given the names aaaaa, aaaab, ...;

and so on.

To sum it up, the dogs numbered 1, 2, \cdots were respectively given the following names:

a, b, ..., z, aa, ab, ..., az, ba, bb, ..., bz, ..., za, zb, ..., zz, aaa, aab, ..., aaz, aba, abb, ..., abz, ..., zzz, aaaa, ...

Now, Roger asks you:

"What is the name for the dog numbered N?"

Constraints

N is an integer.

1 \leq N \leq 1000000000000001

Input

Input is given from Standard Input in the following format:

N

Output

Print the answer to Roger's question as a string consisting of lowercase English letters.

Sample Input 1

2

Sample Output 1

b

Sample Input 2

27

Sample Output 2

aa

Sample Input 3

123456789

Sample Output 3

jjddja

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

/* C - One Quadrillion and One Dalmatians */

#include <stdio.h>
#define ALPH 26

/*
long long ExAlph(int n){
  if(n == 0) return 1;
  else return ALPH * ExAlph(n - 1);
}
*/

int main(void){
  long long N, n;
  long long exAlph[12];
  scanf("%lld", &N);
  for(int i = 0; i < 12; i++){
    if(i == 0){
      exAlph[i] = 1;
    }else{
      exAlph[i] = exAlph[i - 1] * ALPH;
    }
    // printf("%lld\n", exAlph[i]);
  }
  n = N;
  for(int i = 1; ; i++){
    if(n <= exAlph[i]){ // 出力される文字数はiで確定
      n--;
      for(int j = i; j > 0; j--){
        printf("%c", (int) 'a' + n / exAlph[j - 1]);
        n -= n / exAlph[j - 1] * exAlph[j - 1];
      }
      break;
    }else{
      n -= exAlph[i];
    }
  }
  printf("\n");
  return 0;
}