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<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<stack>
#include<queue>
#include<map>
#include<functional>
#include<math.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<(n);i++)
typedef long long ll;
typedef pair<ll, ll> p;
const int INF = 1001001001;
const double PI = acos(-1);

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

	string name;

	while (n >0) {
		int a = n % 26;
		if (a) {
			name += 'a' + a;
		}
		else {
			name += 'z';
		}
		
		n /= 26;
		rep(i, n) {
			if (n == 26 * i + 1) {
				n--;
				break;
			}
		}
	}
	if (n == 1) {
		name += 'a';
	}
	reverse(name.begin(), name.end());
	cout << name << endl;

	return 0;
}