#include <bits/stdc++.h>
bool palindrome(int st, int ed, char *ptr) {
  int i, mark = ed;
  for (i = st; i <= ((ed + st) / 2); i++)
    if (ptr[i] != ptr[mark--]) return false;
  return true;
}
int main() {
  int ed, st, i, k, l, div;
  char name[1005];
  scanf(" %[^\n] %d", name, &k);
  bool status = true;
  l = (strlen(name));
  if (l % k != 0) {
    printf("NO\n");
    return 0;
  }
  st = 0;
  ed = div = ((l / k) - 1);
  for (i = 1; i <= k; i++) {
    if (palindrome(st, ed, name)) {
      st = ed + 1;
      ed = st + div;
    } else {
      status = false;
      break;
    }
  }
  if (status == false)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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>
bool palindrome(int st, int ed, char *ptr) {
  int i, mark = ed;
  for (i = st; i <= ((ed + st) / 2); i++)
    if (ptr[i] != ptr[mark--]) return false;
  return true;
}
int main() {
  int ed, st, i, k, l, div;
  char name[1005];
  scanf(" %[^\n] %d", name, &k);
  bool status = true;
  l = (strlen(name));
  if (l % k != 0) {
    printf("NO\n");
    return 0;
  }
  st = 0;
  ed = div = ((l / k) - 1);
  for (i = 1; i <= k; i++) {
    if (palindrome(st, ed, name)) {
      st = ed + 1;
      ed = st + div;
    } else {
      status = false;
      break;
    }
  }
  if (status == false)
    printf("NO\n");
  else
    printf("YES\n");
  return 0;
}

```