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 <stdio.h>
#include<math.h>
//Run with 'Ctrl + Alt + N'

int pow26(int n){
    int i,out=1;
    for (i=0;i<n;i++){
        out *= 26;
    }
    return out;
}

int main(void){

    int data[12]={},loop,i;
    long long temp,N;
    
    scanf("%lld",&N);
    //printf("%lu\n",N);
    for(i=0; i<12;i++){
        if( pow26(i)<=N && N<pow26(i+1)){
            loop = i;
            break;
        }
    }
    //printf("loop = %d\n",loop);
    temp = N;

    for(i=loop; i>=0; i--){
        data[i]= temp / pow26(i);
        temp = temp % (pow26(i));
        printf("%c",data[i]+96);
        //printf("26^%d = %d\n",i,data[i]);
    }
    return (0);
}