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 <cmath>
#include <iomanip>
#include <stack>
#include <algorithm>
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <stdio.h>
#include <math.h>


using namespace std;

typedef long long ll;

typedef vector<vector<int>> vvi;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<string>> vvs;
typedef vector<ll> vll;

typedef pair<int, int> P;

typedef queue<int> qi;
typedef queue<string> qs;


#define rep(i, n) for(int i=0; i<(n); ++i)
#define repp(i, a, b) for(int i = (a); i <= (b); i++)
#define Rep(i, n) for(int i=n; i >= 0; i--)

#define all(v) v.begin(), v.end()
//sort( all(v) )　などと使える

/*
int main() {
	ll n; cin >> n;

	string c;

	
	while(n){
		ll r = n % 26;
		string s;
		s = (char)((int)r + 96);

		s += c;

		c = s;

		//cout << r + 96 << endl;

		n /= 26;
	}

	cout << c << endl;
	return 0;
}
*/



ll ll_max(int a, ll b) {
	ll A = (ll)a;
	if (A > b) return A;
	else return b;
}

int main() {
	ll n; cin >> n;
	string ans;

	while (n) {
		n--;
		ans = (char)((int)'a' + n % 26) + ans;
		n /= 26;
	}

	cout << ans << endl;
}