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)

#include<bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;

void Main()
{
  ll N; cin >> N;
  ll slen = 0;
  string ans = "";
  ll num = N;
  
  if (N%(ll)26 > 0) {
    ans += (char)(N%(ll)26 + 'a' - 1);
    num -= N%(ll)26;
  } else {
    ans += 'z';
    num -= 26;
  }
  ll p = 1;
  
  while (pow((ll)26,p) <= N) {
    //cout << num << endl;
    if (num/(ll)pow((ll)26,p) > 0) {
      if ((num/(ll)pow((ll)26,p)) % ((ll)26) > 0) 
      {  ans += (char)( (num/(ll)pow((ll)26,p)) % ((ll)26) + 'a' - 1);
        num = num - ((num/(ll)pow((ll)26,p)) % ((ll)26))*(ll)pow((ll)26,p);
      } else {
        ans += 'z';
        num = num - (ll)pow((ll)26,p);
      }
      
    } 
    p++;
  }
  reverse(ans.begin(), ans.end());
  cout << ans << endl;
}

int main()
{
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  Main();
  return 0;
}