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)

/* abc_171_c */

#include<stdio.h>
#include<math.h>

int main()
{
  long long N;
  int t,s,i;
  long long p[20],q[20],a[20],b[20],d[20];
  char c[20];
  b[0]=1;
  d[0]=1;
  for(i=0;i<=10;i++){
    b[i+1]=26*b[i];
    d[i+1]+=b[i+1];
  }
  t=1;
  scanf("%lld",&N);
  for(i=10;i>=0;i--){
    if(N>=d[i]){
      s=i;
      break;
    }
  }
  a[0]=N-1;
  p[0]=a[0]/26;
  q[0]=a[0]%26;
  if(p[0]>0){
    for(i=0;i<20;i++){
      if(p[i]>0){
        t++;
        a[i+1]=p[i]-1;
        p[i+1]=a[i+1]/26;
        q[i+1]=a[i+1]%26;
      } else break;
    }
  }
  for(i=t-1;i>=0;i--){
    c[i]='a'+q[i];
    printf("%c",c[i]);
  }
  printf("\n");
  /*a[s]=N-d[s];
  p[s]=a[s]/26;
  q[s]=a[s]%26;
  if(p[s]>0){
    for(i=s-1;i>=0;i--){
      p[i]=p[i+1]/26;
      q[i]=p[i+1]%26;
    }
  }
  for(i=s;i>=0;i--){
    c[i]='a'+q[i];
    printf("%c",c[i]);
  }
  printf("\n");*/
  return 0;
}