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>
#define mp make_pair
#define fi first
#define se second
#define VAL 100000
#define ll  long long 
#define INF 10000000000000000LL
#define pll pair<long long , long long>
const int MAX=1e6+9;
const ll MOD=998244353;
const int TOT_PRIMES=19;
const int MAX_A=70;
using namespace std;
int a[MAX];
int b[MAX];
multiset<int>st[MAX] , mx;
int main() 
{

	ios_base::sync_with_stdio(false); 
    cin.tie(NULL);  
    ll n;
    cin>>n;
    string ans="";
    while(n){
      ll rem=n%26-1;
      if(rem<0)
        rem=rem+26;
      ans=ans+(char)(rem+'a');
      n=n/((ll)26);
    }
    reverse(ans.begin() , ans.end());
    cout<<ans;

}