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<iomanip>
#include<cmath>
#include<string.h>
#include<algorithm>
#include<vector>
#include<iterator>
#define pb push_back
typedef long long ll;
using namespace std;
bool six(long long n){
	long long s=27;
	for(int i=2;i<13;i++){
		if(s==n) return true;
		else s+=pow(26,i);
	}
	return false;
}
int main()
{	
	ll n;
	char x1,y;
	std::vector<char> v;
	cin>>n;
	while(n>0){
		if(n<27 && n>0){x1=char(96+n);v.pb(x1);n=0;}
		else if(six(n) && n>27){ y='z';v.pb(y);n=n/26-1;}
		else if(n%26==0){ y='z';v.pb(y);n/=26;}
		else{
			x1=char((n%26)+96);
			v.pb(x1);
			n/=26;
		}
	}
	reverse(v.begin(),v.end());
	for(auto& x:v){
		cout<<x;
	}
	cout<<endl;

	return 0;
}