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 <iostream>
#include <algorithm>
#include <bits/stdc++.h>
#include <cmath>
#include <string.h>
//26進数問題
using namespace std;
using ll = long long;

ll exList[]=

{26,702,18278,475254,12356630,321272406,8353082582,217180147158,5646683826134,
146813779479510};
char Alpha[]={"abcdefghijklmnopqrstuvwxyz"};
string result;
ll gcd(ll a, ll b);
int main(){
cin.tie(0);
ll n;
  
cin >>n;
for(int i=0;i<11;i++){
 if(n<=exList[i]){
   if(i==0){
     cout << Alpha[n-1];
     return 0;
   }
   else {
     for(int c=i;c>=0;c--){
       ll temp=int(n/pow(26,c));
	if(temp==0)result+='z';
      else  result+=Alpha[temp-1];
       n-=temp*pow(26,c);
	cout <<"temp:" <<temp;
	cout <<"    result:" <<result <<endl;

     }

    cout <<result;
    return 0;
   }
 }
}

return 0;
}






ll gcd(ll a, ll b){

   if (a%b == 0)
   {
       return(b);
   }
   else
   {
       return(gcd(b, a%b));
   }

}