#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int n, i, j;
  cin >> s;
  cin >> n;
  if (s.length() < n || s.length() % n != 0)
    cout << "NO";
  else {
    int count = n;
    n = s.length() / n;
    int start = 0;
    int end = n - 1;
    bool possible;
    while (count--) {
      possible = true;
      i = start;
      j = end;
      while (i < j) {
        if (s[i] != s[j]) {
          possible = false;
          break;
        }
        i++;
        j--;
      }
      if (!possible) break;
      start += n;
      end += n;
    }
    if (possible)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}

### Prompt

Your task is to create a cpp solution to 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;
  int n, i, j;
  cin >> s;
  cin >> n;
  if (s.length() < n || s.length() % n != 0)
    cout << "NO";
  else {
    int count = n;
    n = s.length() / n;
    int start = 0;
    int end = n - 1;
    bool possible;
    while (count--) {
      possible = true;
      i = start;
      j = end;
      while (i < j) {
        if (s[i] != s[j]) {
          possible = false;
          break;
        }
        i++;
        j--;
      }
      if (!possible) break;
      start += n;
      end += n;
    }
    if (possible)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}

```