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 REP(i, x) for (int i = 0; i < (int)(x); i++)
#define REPS(i, x) for (int i = 1; i <= (int)(x); i++)
#define RREP(i, x) for (int i = ((int)(x)-1); i >= 0; i--)
#define RREPS(i, x) for (int i = ((int)(x)); i > 0; i--)
#define FOR(i, a, b) for (int i = a; i <= (int)(b); i++)
#define FORD(i, a, b) for (int i = a; i >= (int)(b); i--)

#define SIZE_OF_ARRAY(array) (sizeof(array) / sizeof(array[0]))
using Lint = long long;
#define INF 1000000000000
#define MOD 1000000007 //10^9+7:合同式の法

#define ALL(x) (x).begin(), (x).end()
#define MAX(x) *max_element(ALL(x))
#define MIN(x) *min_element(ALL(x))

#define PB push_back
#define MP make_pair
using P=pair<int, int>;
using V=vector<int>;
using M=map<int, int>;
using S=set<int>;

#define COUT(x) cout << (x) << endl
#define SCOUT(x) cout << (x) << " "

#define DUMP(x) cout << #x << " = " << (x) << endl;
#define YES(n) cout << ((n) ? "YES" : "NO") << endl
template <class T>bool chmax(T &a, const T &b){if (a < b){a = b;return 1;}return 0;}
template <class T>bool chmin(T &a, const T &b){if (b < a){a = b;return 1;}return 0;}
/* vector<T>のイテレータはvector<T>::iterator型
vector<int>::iterator it = vc.begin();  // begin()はvectorの最初の要素を指すイテレータを返す。
vector<int>::iterator endIt = vc.end(); // end()はvectorの終端位置を指すイテレータを返す。
while (it != endIt)
{
  cout << *it << '\n'; // イテレータの指す要素は、ポインタのように*演算子でアクセスします。
  it++;                // インクリメントすることで次の要素を指します。
}
}*/
 
int main()
{
  
  int n;
  cin >>n;
  

  
  COUT('a'); 
}