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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

#define REP(i, n) for (ll i = 0; i < (ll)(n); i++)
#define RREP(i, n) for (ll i = n - 1; i >= 0; i--)
#define FOR(i, a, b) for (ll i = a; i <= (ll)(b); i++)
#define RFOR(i, a, b) for (ll i = a; i >= (ll)(b); i--)
#define ALL(x) (x).begin(), (x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
    os << "[ ";
    for (auto a : v) os << a << " ";
    os << "]";
    return os;
}

int main() {
    ll N;
    cin >> N;
    string ans = "";
    while (N > 0) {
        N--;
        ans += 'a' + (N % 26);
        N /= 26;
    }
    reverse(ALL(ans));
    cout << ans << endl;
    return 0;
}