#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int count = 0;
  int prev_i, i, prev_j, j;
  prev_i = i = 0;
  prev_j = j = (s.size() / k) - 1;
  int each = s.size() / k;
  if (s.length() % k != 0) {
    cout << "NO";
    return 0;
  }
  while (i <= j) {
    if (s[i] == s[j]) {
      count++;
      i++;
      j--;
    } else {
      cout << "NO";
      return 0;
    }
    if (count == each / 2 || (each == 1 && count == 1)) {
      if (prev_j == s.size() - 1) {
        cout << "YES";
        return 0;
      }
      i = prev_i + each;
      j = prev_j + each;
      count = 0;
      prev_i = i;
      prev_j = j;
    }
  }
}

### Prompt

Create a solution in cpp for the following problem:

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;
  cin >> s;
  int k;
  cin >> k;
  int count = 0;
  int prev_i, i, prev_j, j;
  prev_i = i = 0;
  prev_j = j = (s.size() / k) - 1;
  int each = s.size() / k;
  if (s.length() % k != 0) {
    cout << "NO";
    return 0;
  }
  while (i <= j) {
    if (s[i] == s[j]) {
      count++;
      i++;
      j--;
    } else {
      cout << "NO";
      return 0;
    }
    if (count == each / 2 || (each == 1 && count == 1)) {
      if (prev_j == s.size() - 1) {
        cout << "YES";
        return 0;
      }
      i = prev_i + each;
      j = prev_j + each;
      count = 0;
      prev_i = i;
      prev_j = j;
    }
  }
}

```