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>

int main(){
    long int i;
    int j,k=0,l;
    char a[100];
    if(scanf("%ld",&i)!=1)
        i=1;
    while(i!=0){
        j=i%26;
        i=i/26;
        switch (j){
        case 0:
            a[k]='z';
            break;
        case 1:
            a[k]='a';
            break;
        case 2:
            a[k]='b';
            break;
        case 3:
            a[k]='c';
            break;
        case 4:
            a[k]='d';
            break;
        case 5:
            a[k]='e';
            break;
        case 6:
            a[k]='f';
            break;
        case 7:
            a[k]='g';
            break;
        case 8:
            a[k]='h';
            break;
        case 9:
            a[k]='i';
            break;
        case 10:
            a[k]='j';
            break;
        case 11:
            a[k]='k';
            break;
        case 12:
            a[k]='l';
            break;
        case 13:
            a[k]='m';
            break;
        case 14:
            a[k]='n';
            break;
        case 15:
            a[k]='o';
            break;
        case 16:
            a[k]='p';
            break;
        case 17:
            a[k]='q';
            break;
        case 18:
            a[k]='r';
            break;
        case 19:
            a[k]='s';
            break;
        case 20:
            a[k]='t';
            break;
        case 21:
            a[k]='u';
            break;
        case 22:
            a[k]='v';
            break;
        case 23:
            a[k]='w';
            break;
        case 24:
            a[k]='x';
            break;
        default:
            a[k]='y';
            break;
        }

        k++;

    }
    for(l=k;l>=0;l--){
        printf("%c",a[l]);
    }
    printf("\n");
    return 0;
}