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;

typedef long long ll;
typedef pair<int, int> pii;
#define FOR(i, n) for(int (i)=0; (i)<(n); (i)++)
#define FOR1(i, n) for(int (i)=1; (i)<=(n); (i)++)
#define FORI(i, n) for(int (i)=n-1; (i)>=0; (i)--)

int main(int argc, char** argv) {
    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cout << setprecision(15);
    if (argc == 2 && atoi(argv[1]) == 123456789) freopen("d:\\code\\cpp\\contests\\stdin", "r", stdin);

    ll n;
    cin >> n;

    string res;
    ll q = 1;
    while(q*26 < n){
        q *= 26;
    }
    while(q) {
        res += 'a' + n / q - 1;
        n %= q;
        q /= 26;
    }
    cout << res << endl;

    if (argc == 2 && atoi(argv[1]) == 123456789) cout << clock()*1.0/CLOCKS_PER_SEC << " sec\n";
    return 0;
}