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 (Clang 10.0.0) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(void)
{
    long long n;
    scanf("%lld", &n);

    long long q, r;
    char ans[1000];
    int i,j=1;
    long long judge=26,tmp;
    while(1){
        if(judge>=n){
            break;
        }else{
            judge *= 26;
            judge += 26;
            /*一つずらす*/
            j++;
            /*jは桁数*/
        }
    }
    for(i=0;i<j;i++){
        ans[i]='a';
        /*j桁をaで初期化*/
    }
    judge -=26;
    judge /=26;
    /*一つもどす*/
    n-=judge;
    /*j桁において何番目かで数える*/
    i=1;
    tmp=j;
    /*桁数を保存*/
    while (1)
    {
        r = n%26;
        if(r!=0){
            ans[j - 1] += -1 + r;
            j--;
        }else{
            ans[j-1] = 'z';
            j--;
        }
        n/=26;
    
        if(n==0){
            break;
        }
    }
    for (i=0;i<tmp;i++)
    {
        putchar(ans[i]);
    }
    putchar('\n');
    return 0;
}