#include <bits/stdc++.h>
int main() {
  char message[1005];
  scanf("%s", message);
  int k, n, i, a, b, j;
  scanf("%d", &k);
  int l = strlen(message);
  if (l % k == 0) {
    n = l / k;
    a = 0;
    b = a + n - 1;
    for (i = 1; i <= k; i++) {
      for (j = a; j <= b; j++) {
        if (message[j] != message[a + b - j]) {
          printf("NO\n");
          return 0;
        }
      }
      a = b + 1;
      b = a + n - 1;
    }
    printf("YES\n");
    return 0;
  } else {
    printf("NO\n");
    return 0;
  }
}

### Prompt

In Cpp, your task is to solve 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>
int main() {
  char message[1005];
  scanf("%s", message);
  int k, n, i, a, b, j;
  scanf("%d", &k);
  int l = strlen(message);
  if (l % k == 0) {
    n = l / k;
    a = 0;
    b = a + n - 1;
    for (i = 1; i <= k; i++) {
      for (j = a; j <= b; j++) {
        if (message[j] != message[a + b - j]) {
          printf("NO\n");
          return 0;
        }
      }
      a = b + 1;
      b = a + n - 1;
    }
    printf("YES\n");
    return 0;
  } else {
    printf("NO\n");
    return 0;
  }
}

```