#include <bits/stdc++.h>
using namespace std;
int main() {
  char a[1000], b[1000];
  cin >> a;
  int k, flag = 0;
  cin >> k;
  int l = strlen(a);
  int x = l / k;
  if (x * k != l) {
    cout << "NO";
    return 0;
  } else {
    for (int i = 0; i < l; i = i + x) {
      for (int j = i + x - 1, q = 0; j >= i; j--, q++) {
        b[q] = a[j];
      }
      for (int j = i, q = 0; j < i + x; j++, q++) {
        if (b[q] != a[j]) {
          flag = 1;
          break;
        }
      }
    }
  }
  if (flag == 1)
    cout << "NO";
  else
    cout << "YES";
}

### 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() {
  char a[1000], b[1000];
  cin >> a;
  int k, flag = 0;
  cin >> k;
  int l = strlen(a);
  int x = l / k;
  if (x * k != l) {
    cout << "NO";
    return 0;
  } else {
    for (int i = 0; i < l; i = i + x) {
      for (int j = i + x - 1, q = 0; j >= i; j--, q++) {
        b[q] = a[j];
      }
      for (int j = i, q = 0; j < i + x; j++, q++) {
        if (b[q] != a[j]) {
          flag = 1;
          break;
        }
      }
    }
  }
  if (flag == 1)
    cout << "NO";
  else
    cout << "YES";
}

```