#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, stemp, stemp2;
  bool bflag = true;
  int n, cont = 0;
  cin >> s;
  cin >> n;
  if (s.length() == 1 && n == 1) {
    cout << "YES" << endl;
    return 0;
  }
  if (s.length() % n != 0) bflag = false;
  for (int i = 0; i < n; i++) {
    if (bflag) {
      stemp = s.substr(i + cont, s.length() / n);
      cont += s.length() / n - 1;
      stemp2 = stemp;
      for (int j = 0; j < stemp.length(); j++) {
        stemp2[j] = stemp[stemp.length() - 1 - j];
      }
      if (stemp == stemp2)
        bflag = true;
      else
        bflag = false;
    }
  }
  if (bflag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

While Mike was walking in the subway, all the stuff in his back-bag dropped on the ground. There were several fax messages among them. He concatenated these strings in some order and now he has string s.

<image>

He is not sure if this is his own back-bag or someone else's. He remembered that there were exactly k messages in his own bag, each was a palindrome string and all those strings had the same length.

He asked you to help him and tell him if he has worn his own back-bag. Check if the given string s is a concatenation of k palindromes of the same length.

Input

The first line of input contains string s containing lowercase English letters (1 ≤ |s| ≤ 1000).

The second line contains integer k (1 ≤ k ≤ 1000).

Output

Print "YES"(without quotes) if he has worn his own back-bag or "NO"(without quotes) otherwise.

Examples

Input

saba
2


Output

NO


Input

saddastavvat
2


Output

YES

Note

Palindrome is a string reading the same forward and backward.

In the second sample, the faxes in his back-bag can be "saddas" and "tavvat".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  string s, stemp, stemp2;
  bool bflag = true;
  int n, cont = 0;
  cin >> s;
  cin >> n;
  if (s.length() == 1 && n == 1) {
    cout << "YES" << endl;
    return 0;
  }
  if (s.length() % n != 0) bflag = false;
  for (int i = 0; i < n; i++) {
    if (bflag) {
      stemp = s.substr(i + cont, s.length() / n);
      cont += s.length() / n - 1;
      stemp2 = stemp;
      for (int j = 0; j < stemp.length(); j++) {
        stemp2[j] = stemp[stemp.length() - 1 - j];
      }
      if (stemp == stemp2)
        bflag = true;
      else
        bflag = false;
    }
  }
  if (bflag)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```