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 <algorithm>
#include <utility>
#include <map>
#include <vector>
#include <cstring>
#include <climits>
#include <queue>
#include <cmath>
#include <iomanip>
#include <set>
#include <bitset>

using namespace std;

typedef long long LL;
typedef pair<int, int> pii;

int main() {
	ios::sync_with_stdio(false);
	LL N;
	cin >> N;
	LL tmp = N;
	LL p = 1;
	while (true) {
		if (N > (LL)pow(26LL, p)) {
			N -= (LL)pow(26LL, p);
			p++;
		} else {
			break;
		}
	}
	N--;
	string ans = "";
	for (LL i = p; i >= 1; i--) {
		LL cur = (LL)pow(26LL, i - 1);
		int num = N / cur;
		ans += (char)('a' + num);
		N -= (num) * cur;
	}
	cout << ans << endl;
}