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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <queue>
#include <cmath>
#include <climits>
#include <iomanip>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
int main(){
  ll n = 0;
  cin >> n;
  //string res;
  //cout << (ll)pow(26,12) << endl;
  vector<char> res(100);
  for(int i = 0;i < 100;i++)res[i] = '(';
  for(int i = 12;i >= 0;i--){
    if(n >= pow(26,i)){
      ll now = 0;
      now = n / pow(26,i);
      n -= now * pow(26,i);
      //cout << i << " " << now << endl;
      res[i] = (char)('a' + now - 1);
      //cout << i << " " << res[i] << endl;
    }
  }
  string name;
  for(int i = 12;i >= 0;i--){
    bool in = false;
    if(i != 0){
    if(res[i] != '(' && res[i - 1] == '('){
      //cout << i << endl;
      name += 'z';
      i--;
      in = true;
    }
    }
    if(res[i] == '(')continue;
    else if(!in)name += res[i];
  }
  //reverse(res.b
  cout << name << endl;
}