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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
typedef long long ll;
#define _GLIBCXX_DEBUG
typedef vector<int> vec;
typedef vector<ll> lvec;
typedef vector<char> cvec;
typedef vector<double> dvec;
typedef pair<ll, ll> LP;
typedef tuple<ll, ll, ll> LT;
typedef pair<int, int> P;
typedef tuple<int, int, int> T;
#define fs first
#define sc second
string f(ll n){
  if(n==0) return ""; //再帰関数の終り
  n--; //'a'=0なのでnも"0から"の基準に合わせる
  return f(n/26)+(char)(n%26); 
  //それぞれ26進数に近いような似たような動きがあるため再帰関数をつくる
  //1の桁はn%26,2桁目以降は26で割ったあとn/26。
}
int main()	{
  ll n;
  cin>>n;
  cout<<f(n)<<endl;
}