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 <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <cstdint>
#include <numeric>
#define INF 1000000000
#define LLINF 2000000000000000
#define MOD 1000000007
using namespace std;
#define FOR(i,a,b) for(int i=a;i<b;i++)
#define LOOP(i,N) for(int i=0;i<N;i++)
#define LOOP1(i,N) for(int i=1;i<=N;i++)
typedef pair<int,int> P;
typedef pair<int,pair<int,int> > PP;
#define int long long

signed main(){
  int n;
  cin >> n;
  string ans(' ',10);
  int cnt = 0;
  while(1){
    n-=1;
    ans[cnt] ='a' + n%26;
    n/=26;
    cnt++;
    if(n==0)break;
  }
  LOOP(i,cnt){
    cout << ans[cnt-i-1];
  }
  // cout << ans << endl;
}