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;
using pii = pair<int, int>;

#define pb push_back
#define mp make_pair
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define x first
#define y second

const int MOD = 1e9 + 7;

void setIO() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
}

//ctrl q is comment
//ctrl b is run
//ctrl s is compile
//shift r is search & replace
//shift d is delete a line

bool comp(pii a, pii b){
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}
int main(){
    setIO();
    ll n; cin >> n;
    int p26 = 0;
    ll mult = 26;
    int temp = n;
    while(temp > mult){
        temp -= mult;
        mult *= 26;
        ++p26;
    }
    mult = 1;
    for(int i = 0; i < p26; ++i){
        mult *= 26;
    }
    string ans = "";
    for(int i = p26; i >= 0; --i){
        if(n % mult == 0 && mult != 1){
            ans += (char)(n / mult - 2 + 'a');
            n %= mult;
            n += mult;
        }
        else{
            ans += (char)(n / mult - 1 + 'a');
            n %= mult;
        }
        mult /= 26;
    }
    cout << ans << "\n";
}