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 int long long
#define ld long double
#define pii pair<int,int>
#define pb push_back
#define pf push_front
#define pof pop_front
#define mp make_pair
#define x first
#define y second

signed main(){
	ios::sync_with_stdio(false);
   	cin.tie(0);

	int n;
	cin>>n;
	if (n%26==0){
		int ret=0;
		while (n>=26){
			ret++;
			n/=26;
		}
		for (int i=0; i<ret; i++)
			cout<<"z";
		cout<<endl;
		return 0;
	}
	string s="abcdefghijklmnopqrstuvwxyz";
	vector<char> ans;
	while (n>=26){
		int x=n%26;
		if (x==0) x=26;
		ans.pb(s[x-1]);
		n/=26;
	}
	reverse(ans.begin(),ans.end());
	cout<<s[n-1];
	for (int i=0; i<ans.size(); i++)
		cout<<ans[i];
	cout<<endl;
	return 0;
}