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 <algorithm>
#include <cstdio>
#include <iostream>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stack>
#include <vector>
using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define Sort(Array) sort(Array.begin(), Array.end())
#define Reverse(a) reverse(a.begin(), a.end())
#define out(ans) cout << ans << endl;
const int MOD = 1000000007;
const int INF = 2147483647;
const ld PI = 3.14159265358979323846;

//------------↓------------- M -------------- E ---------------- M --------------- O ---------------↓--------------//
// コンパイル 
// g++ -std=c++1z
//
// -------型変換--------
// int を string に変換
// string s = to_string(n);
// string を int に変換
// int n = stoi(s);
//
// -------二分探索---------
// k以上の値が最初に現れる時のイテレータ
// lower_bound(data.begin(), data.end(), k)
// kより大きい値が最初の現れる時のイテレータ O(logN)
// upper_bound(data.begin(), data.end(), k)
// kがdataに存在するかをtrue or falseで返す O(logN)
// binary_search(data.begin(), data.end(), k)
// 
//
//
//
//
//
// 
//------------↑------------- M -------------- E ---------------- M --------------- O ---------------↑--------------//






int main() {

    ll N; cin >> N;
    string ans = "";
    while (N > 0) {
        int m = N % 26;
        if (m == 0) {
            m = 26;
        }
        ans += 'a' - 1 + m;
        N /= 26;
    }
    Reverse(ans);
    out(ans);


    return 0;
}