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: Runtime Error)

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

typedef long long ll;

ll index_sum(int x){
    ll sum=0;
    for(int i=1; i<=x; i++){
        sum += pow(26, i);
    }
    return sum;
}

int main(void){
    int num, i, j, k;
    char alphabet, str[10];

    scanf("%d", &num);

    for(i=1; i<10; i++){
        if(num > index_sum(i-1) && num <= index_sum(i)) break;
    }

    k = i;
    for(i=k; i>1; i--){
        for(j=0; j<=25; j++){
            if(num > j * pow(26, i-1) + index_sum(i-1) && num <= (j+1) * pow(26, i-1) + index_sum(i-1)){
                alphabet =(char)(97 + j);
                strcat(str, &alphabet);
                num -= (j+1) * pow(26, i-1);
                break;
            }
        }
    }
    for(j=1; j<=26; j++){
        if(num == j){
            alphabet =(char)(97 + j - 1);
            strcat(str, &alphabet);
            break;
        }
    }


    printf("%s\n", str);

    return 0;
}