#include <bits/stdc++.h>
using namespace std;
int compare(const void *a, const void *b) { return (*(int *)a - *(int *)b); }
int main() {
  char str[1002];
  int n, i, j;
  cin >> str;
  cin >> n;
  int x = strlen(str) / n;
  int l = 0, r = x - 1, flag = 0;
  if (strlen(str) % n == 0) {
    for (i = 0; i < n; i++) {
      int temp = r;
      for (j = l; j <= r; j++) {
        if (str[j] != str[temp]) {
          flag = 1;
          break;
        }
        temp--;
      }
      l = l + x;
      r = r + x;
      if (flag == 1) break;
    }
  } else {
    flag = 1;
  }
  if (flag == 1)
    cout << "NO";
  else
    cout << "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 compare(const void *a, const void *b) { return (*(int *)a - *(int *)b); }
int main() {
  char str[1002];
  int n, i, j;
  cin >> str;
  cin >> n;
  int x = strlen(str) / n;
  int l = 0, r = x - 1, flag = 0;
  if (strlen(str) % n == 0) {
    for (i = 0; i < n; i++) {
      int temp = r;
      for (j = l; j <= r; j++) {
        if (str[j] != str[temp]) {
          flag = 1;
          break;
        }
        temp--;
      }
      l = l + x;
      r = r + x;
      if (flag == 1) break;
    }
  } else {
    flag = 1;
  }
  if (flag == 1)
    cout << "NO";
  else
    cout << "YES";
  return 0;
}

```