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(void){
    //1,000,000,000,000,001
    //26,26^2,26^3
    long n,l,k = 0,z = 1;
    int x,keta;
    int syutu[20] = {0};//わからんけどこれくらいで
    scanf("%ld",&n);
    //n -= 1;
    l = n;
    for(keta = 1;;keta++){
        z *= 26;
        if(l <= z){
            break;
        }
        k += z;
        l -= z;
    }
    //想定通りprintf("%d\n",keta);
    //多分大丈夫printf("%ld\n",l);
    l = n;
    //z /= 26;
    l -= k;
    z = 26;
    for(x = 0;x < keta;x++){
        //z *= 26;
        syutu[x] = (int)l % z;
        l /= z;
    }
    for(keta -= 1;keta >= 0;keta--){
        if(syutu[keta] == 0){
            syutu[keta] += 26;
        }
        printf("%c",syutu[keta] + 'a' - 1);
    }
    
    
    /*while(1){
        if(l < 26){
            break;
        }
        l = l / 26;
        keta++;
    }
    n += 1;
    printf("%d",keta);*/
    //純粋なn進数じゃない
    
    /*for(kai = keta;kai >= 1;kai--){
        k = 1;
        for(x = 1;x < kai;x++){
            k *= 26;
        }
        l = (n - n % k) / k;
        n -= l * k;
        z = (int)l;
        if(z == 0)z += 26;
        printf("\nz = %d\n",z);
        printf("%c",z + 'a' - 1);
        
    }*/
    
    printf("\n");
    return 0;
}