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>
long long g(int a, int n){
    if(n == 0)
        return 1;
    return a * g(a,n - 1);
}
void f(long long n){
    if(n <= 26){
        printf("%c", n + 'a' - 1);
        return ;
    }
    int p;
    for(int i = 1; ;i++){
        if(g(26,i) <= n && n < g(26,i + 1)){
            p = i;
            break;
        }
        else if(g(26,i) <= n && n / 26 < g(26, i)){
            p = i;
            break;
        }
    }
    p++;
    for(int i = p - 1; i > 0; i--){
        long long q = g(26,i);
        int r = n / q;
        printf("%c", r + 'a' - 1);
        n %= q;
    }
    if(n % 26 == 0)
        puts("z");
    else 
        printf("%c\n",n % 26 + 'a' - 1);
}
int main(){
    long long n;
    scanf("%lld",&n);
    f(n);
    return 0;
}