#include <bits/stdc++.h>
using namespace std;
int main() {
  char s[1005];
  int k, f = 0, b;
  scanf("%s", s);
  scanf("%d", &k);
  int t = strlen(s);
  if (t % k != 0) {
    printf("NO");
  } else {
    int x = t / k;
    b = 0;
    while (b < t) {
      for (int i = b, j = b + x - 1; i < b + x && j >= b; i++, j--) {
        if (s[j] == s[i])
          f = 1;
        else {
          f = 0;
          break;
        }
      }
      if (f == 0) break;
      b += x;
    }
    if (f == 1)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

### Prompt

Please formulate 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 s[1005];
  int k, f = 0, b;
  scanf("%s", s);
  scanf("%d", &k);
  int t = strlen(s);
  if (t % k != 0) {
    printf("NO");
  } else {
    int x = t / k;
    b = 0;
    while (b < t) {
      for (int i = b, j = b + x - 1; i < b + x && j >= b; i++, j--) {
        if (s[j] == s[i])
          f = 1;
        else {
          f = 0;
          break;
        }
      }
      if (f == 0) break;
      b += x;
    }
    if (f == 1)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

```