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;

#define IOS                       \
    ios_base::sync_with_stdio(0); \
    cin.tie(NULL);                \
    cout.tie(NULL);
#define pb push_back
#define mp make_pair
#define Debug(x) cout << #x " = " << (x) << endl
#define SORT(a) sort(a.begin(), a.end())
#define SORTR(a) sort(a.rbegin(), a.rend())

int main()
{
    IOS;

    int n;
    cin >> n;

    if (n > 26)
    {
        vector<char> s;
        while (n >= 25)
        {
            int x = (n % 26);
            n -= x;
            n /= 26;

            s.pb(char(x + 96));
        }
        s.pb(char(n + 96));

        for (int i = s.size() - 1; i >= 0; i--)
        {
            cout << s[i];
        }
    }
    else
    {
        cout << char(n + 96) << endl;
    }
    return 0;
}