#include <bits/stdc++.h>
int main() {
  char a[10000];
  float f, total;
  int i, j, n, h, m = 0, M = 0, x, h1, q;
  for (i = 0; i < 10000; i++) {
    scanf("%c", &a[i]);
    if (a[i] == '\n') break;
  }
  scanf(" %d", &n);
  total = i;
  h = total / n;
  f = total / n;
  if (h != f) printf("NO");
  h1 = h - 1;
  if (h == f) {
    for (x = 0; x < n; x++) {
      if (h != f) break;
      for (i = 0, j = h - 1; i < h, j >= 0; i++, j--) {
        if (a[i] == a[j]) M++;
      }
      for (q = 0; q < h; q++) {
        h1++;
        a[q] = a[h1];
      }
    }
    if (M == n * h)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

### Prompt

Your challenge is to write 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 main() {
  char a[10000];
  float f, total;
  int i, j, n, h, m = 0, M = 0, x, h1, q;
  for (i = 0; i < 10000; i++) {
    scanf("%c", &a[i]);
    if (a[i] == '\n') break;
  }
  scanf(" %d", &n);
  total = i;
  h = total / n;
  f = total / n;
  if (h != f) printf("NO");
  h1 = h - 1;
  if (h == f) {
    for (x = 0; x < n; x++) {
      if (h != f) break;
      for (i = 0, j = h - 1; i < h, j >= 0; i++, j--) {
        if (a[i] == a[j]) M++;
      }
      for (q = 0; q < h; q++) {
        h1++;
        a[q] = a[h1];
      }
    }
    if (M == n * h)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

```