#include <bits/stdc++.h>
char s[1010];
char ss[10] = "gxvug";
int main() {
  scanf("%s", s);
  int k;
  scanf("%d", &k);
  int l = strlen(s);
  if (k > l || l % k) {
    printf("NO\n");
    return 0;
  }
  k = l / k;
  for (int i = 0; i < l; i += k) {
    int j = i + k - 1;
    int t = i;
    while (t <= j) {
      if (s[t] == s[j]) {
        t++;
        j--;
      } else {
        printf("NO\n");
        return 0;
      }
    }
  }
  printf("YES\n");
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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>
char s[1010];
char ss[10] = "gxvug";
int main() {
  scanf("%s", s);
  int k;
  scanf("%d", &k);
  int l = strlen(s);
  if (k > l || l % k) {
    printf("NO\n");
    return 0;
  }
  k = l / k;
  for (int i = 0; i < l; i += k) {
    int j = i + k - 1;
    int t = i;
    while (t <= j) {
      if (s[t] == s[j]) {
        t++;
        j--;
      } else {
        printf("NO\n");
        return 0;
      }
    }
  }
  printf("YES\n");
  return 0;
}

```