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)

#pragma region Macros
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

#define REP(i, n) for(int i=0; i<(n); i++)
#define REPS(i, n) for(int i=0; i<=(n); i++)
#define FOR(i, a, b) for(int i=a; i<(b); i++)
#define FORS(i, a, b) for(int i=a; i<=(b); i++)
// for container x
#define ALL(v) (v).begin(),(v).end()
#define RALL(v) (v).rbegin(),(v).rend()
#define SIZE(v) ((ll)(v).size())
#define MAX(v) *max_element(ALL(v))
#define MIN(v) *min_element(ALL(v))
#define UNIQUE(v) v.erase(unique(ALL(v)), v.end())

struct cww{cww(){ios::sync_with_stdio(false);cin.tie(0);}}star;
#pragma endregion

int main(){
    ll n;
    cin >> n;
    n=n-1;
    int remainder;
    string ans="";
    while (n>=0) {
        remainder = n % 26;
        // ans = remainder+'a';
        ans.insert(ans.begin(), remainder+'a');
        n /= 26;
        n -= 1;
    }
    cout << ans << endl;
    return 0;
}