#include <bits/stdc++.h>
using namespace std;
int t, i, j, k, ans;
char str[1001];
int main() {
  int flag = 0;
  scanf("%s", str);
  cin >> k;
  int len = strlen(str);
  if (len % k != 0)
    flag = 1;
  else
    t = len / k;
  if (flag != 1) {
    for (i = 0; i < len; i += t) {
      int k = i;
      j = i + t - 1;
      while (i < j) {
        if (str[k] == str[j]) {
          k++;
          j--;
        } else {
          flag = 1;
          break;
        }
      }
      if (flag == 1) break;
    }
  }
  if (flag)
    printf("NO");
  else
    printf("YES");
  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>
using namespace std;
int t, i, j, k, ans;
char str[1001];
int main() {
  int flag = 0;
  scanf("%s", str);
  cin >> k;
  int len = strlen(str);
  if (len % k != 0)
    flag = 1;
  else
    t = len / k;
  if (flag != 1) {
    for (i = 0; i < len; i += t) {
      int k = i;
      j = i + t - 1;
      while (i < j) {
        if (str[k] == str[j]) {
          k++;
          j--;
        } else {
          flag = 1;
          break;
        }
      }
      if (flag == 1) break;
    }
  }
  if (flag)
    printf("NO");
  else
    printf("YES");
  return 0;
}

```