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<stdio.h>
#include<math.h>
#include<string.h>
#include<string>
#include<stack>
#include<map>
#include<set>
#include<queue>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
#define PI acos(-1)
#define pb push_back
#define all(x) x.begin(),x.end()
#define INF 0x3f3f3f3f
#define dd(x) cout<<#x<<" = "<<x<<","
#define de(x) cout<<#x<<" = "<<x<<"\n"
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int N=1e6+5;
int a[N];
char ans[N];
int main()
{
	ll n;
	scanf("%lld",&n);
	ll sum=0;
	int len=1;
	ll i=1;
	for(;;i*=26){
		if(sum+i*26>=n){
			break;
		}
		sum+=i*26;
		len++;
	}
	n-=sum;
	n--;
	for(int i=1;i<=len;i++){
		ans[i]='a';
		ans[i]+=n%26;
		n/=26;
	}
	for(int i=len;i>=1;i--){
		printf("%c",ans[i]);
	}
}