#include <bits/stdc++.h>
int len, i, j, n, x;
char st[1010];
int main() {
  scanf("%s", st);
  len = strlen(st);
  scanf("%d", &n);
  if (len % n != 0) {
    printf("NO\n");
    return 0;
  }
  x = len / n;
  for (i = 0; i < n; i++) {
    for (j = i * x; j < (i + 1) * x; j++) {
      if (st[j] != st[(i + 1) * x - 1 - j + i * x]) {
        printf("NO\n");
        return 0;
      }
    }
  }
  printf("YES\n");
}

### Prompt

Generate 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>
int len, i, j, n, x;
char st[1010];
int main() {
  scanf("%s", st);
  len = strlen(st);
  scanf("%d", &n);
  if (len % n != 0) {
    printf("NO\n");
    return 0;
  }
  x = len / n;
  for (i = 0; i < n; i++) {
    for (j = i * x; j < (i + 1) * x; j++) {
      if (st[j] != st[(i + 1) * x - 1 - j + i * x]) {
        printf("NO\n");
        return 0;
      }
    }
  }
  printf("YES\n");
}

```