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;
using ll = long long;
#define rep(i, n) for(i = 0; i < n; i++)
//2D::rep(i,j)rep(j,j)cin>>bin[i][j];
//vector<int> a(n)
#define Graph vector<vector<int>>
#define A1 cout<<"No"<<endl
#define A2 cout<<"Yes"<<endl
#define A3 cout<<ans
#define AA(i) cout<<i<<endl;
const int U=2e5+5;
const ll MOD=1000000007;
const int INF = 1001001001;

int main(){
    ll N; cin >> N;
    char s = 'a';
    string S;
    ll i=26,j=1;
    while(N > i){
        N -= i;
        i *= 26;
        j++;
    }
    if(N%26==0){
        s = 'z';
    }
    else{
            s+=N%26-1;
        }
    S=s;
    if(N==26) N=0;
    N/=26;
    while(j>1){
        s = 'a';
        if(N==0){}
        else if(N%26==0){
            s = 'z';
        }
        else{
        s += N%26;
        }
        S = s + S;
        N /= 26;
        if(N==1) N==0;
        j--;
    }
    AA(S);
    return 0;
}