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 int N;
char name[13];

int main(){
    
    int quot=0,rema=0;

    scanf("%lld",&N);

    int i=0,last=0;
    for(i=0;i<13;i++){
	quot=N/26;
	rema=N-N/26;

	if(rema==0)
	    name[i]='z';
	else if(rema==1)
	    name[i]='a';
	else if(rema==2)
	    name[i]='b';
	else if(rema==3)
	    name[i]='c';
	else if(rema==4)
	    name[i]='d';
	else if(rema==5)
	    name[i]='e';
	else if(rema==6)
	    name[i]='f';
	else if(rema==7)
	    name[i]='g';
	else if(rema==8)
	    name[i]='h';
	else if(rema==9)
	    name[i]='i';
	else if(rema==10)
	    name[i]='j';
	else if(rema==11)
	    name[i]='k';
	else if(rema==12)
	    name[i]='l';
	else if(rema==13)
	    name[i]='m';
	else if(rema==14)
	    name[i]='n';
	else if(rema==15)
	    name[i]='o';
	else if(rema==16)
	    name[i]='p';
	else if(rema==17)
	    name[i]='q';
	else if(rema==18)
	    name[i]='r';
	else if(rema==19)
	    name[i]='s';
	else if(rema==20)
	    name[i]='t';
	else if(rema==21)
	    name[i]='u';
	else if(rema==22)
	    name[i]='v';
	else if(rema==23)
	    name[i]='w';
	else if(rema==24)
	    name[i]='x';
	else
	    name[i]='y';
	
	if(quot==0)
	    break;
	else if(quot==1 && rema==0)
	    break;

	N=quot;
    }

    last=i;

    for(i=last;i>=0;i--)
	printf("%c",name[i]);

    printf("\n");

    return 0;
}