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: Accepted)

int main(void){
  
    long long int input;
    int cnt, name[100] = {0};
    int memo;
  
    scanf("%lld", &input);
    //memo = input;
  
    //--;
    /*for(int j = 1; j <= memo; j++){  
        input = j;
        cnt = 0;*/
        
    for(int i = 0; ; i++){
        if(input == 0){
            break;
        } else {
	        name[i] = input % 26;
	        input = input / 26;
	        //printf("%d %lld\n", name[i], input);
	        cnt++;
        }
    }
    
    for(int i = 0; i < cnt; i++){
        if(name[i] < 1){
            name[i] += 26;
            name[i + 1]--;
            if(name[cnt - 1] == 0){
                cnt--;
            }
        }
    }
    
    //name[0];
    cnt--;
    
    //printf("%d: ", j);                                 
    for(int i = 0; i <= cnt; i++){
        //printf("%lld", name[cnt - i]);
        printf("%c", name[cnt - i] + 'a' - 1);
    }
    printf("\n");
    
    //}
  
    return 0;
}