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: Wrong Answer)

#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define ll long long
// 1,000,000,000,000,001
int main(void)
{
  ll n, mo;
  int exp = 1;
  ll exmax = 26;
  ll max[12] = {0};
  ll mdif[12] = {1};
  //char c[27];// c[1] = 97,,,,c[26] = 122
  scanf("%lld", &n);
  while (1)
  {

    max[exp] = max[exp - 1] + mdif[exp - 1] * 26;
    mdif[exp] = mdif[exp - 1] * 26;
    //printf("%lld %lld\n", max[exp], mdif[exp]);
    if (n <= max[exp] || (exp == 11))
      break;
    exp++;
  } // ともに1 ~ 有効
  //max[exp-1] ~ max[exp]までに答えがある。その間にはmdif[exp]個の要素がある
  //また現れる文字の数はexp個
  while (exp > 1)
  {
    mo = n / mdif[exp - 1];
    printf("%c", (char)mo + 96);
    //printf("%lld   ", mo);
    n -= mo * mdif[exp - 1];
    exp--;
  }
  //printf("\n%lld\n", n);
  printf("%c\n", (char)n + 96);
  return 0;
}