#include <bits/stdc++.h>
using namespace std;
int main() {
  char input[1002];
  int i, k, l, len, j, res, m;
  scanf("%s", input);
  scanf("%d", &k);
  len = strlen(input);
  if (len == k)
    printf("YES\n");
  else if (len % k != 0)
    printf("NO\n");
  else {
    res = 0;
    for (i = 0; input[i] != '\0'; i = i + (len / k)) {
      for (j = i, l = i + (len / k) - 1, m = 0; m < (len / k) / 2;
           m++, j++, l--) {
        if (input[j] != input[l]) {
          res = 1;
          break;
        }
      }
      if (res == 1) break;
    }
    if (res == 1)
      printf("NO\n");
    else
      printf("YES\n");
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 input[1002];
  int i, k, l, len, j, res, m;
  scanf("%s", input);
  scanf("%d", &k);
  len = strlen(input);
  if (len == k)
    printf("YES\n");
  else if (len % k != 0)
    printf("NO\n");
  else {
    res = 0;
    for (i = 0; input[i] != '\0'; i = i + (len / k)) {
      for (j = i, l = i + (len / k) - 1, m = 0; m < (len / k) / 2;
           m++, j++, l--) {
        if (input[j] != input[l]) {
          res = 1;
          break;
        }
      }
      if (res == 1) break;
    }
    if (res == 1)
      printf("NO\n");
    else
      printf("YES\n");
  }
  return 0;
}

```