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)

// by azidoazideazide(C2N14)
#include<iostream>
#include<cmath>
#include<algorithm>
#include<vector>
#include<utility>
using namespace std;
typedef long long ll;
#define pb push_back
#define fi first
#define se second

ll ck[12],r[12];
string s;

int main(){
	ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
	
	for(ll i=0,x=1;i<11;i++,x*=26){
		r[i]=ck[i]=x;
		if(i>0) r[i]+=r[i-1];
		
	} 
	
//	for(int i=0;i<11;i++) cout<<i<<" "<<ck[i]<<"    "<<r[i]<<endl;//deb
	
	ll n;
	cin>>n;
	
	ll pos=10;
	while(n>0&&pos>=0){
		if(n>=r[pos]){
			int no=n/ck[pos];
			if(no){
//				cout<<no<<endl;//deb
				s+=('a'+no-1);
//				cout<<s<<endl;//deb
				n-=no*ck[pos];
			}
		}
		pos--;
	}
	cout<<s<<endl;
}