#include <bits/stdc++.h>
using namespace std;
int main() {
  string str;
  cin >> str;
  int n, flag, len;
  cin >> n;
  if (str.length() % n != 0 || str.length() < n)
    cout << "NO\n";
  else {
    len = str.length() / n;
    for (int i = 0; i < n; i++) {
      int last = (i + 1) * len - 1;
      flag = 1;
      for (int j = i * len; j < last; j++) {
        if (str[j] == str[last])
          last--;
        else {
          flag = 0;
          break;
        }
      }
      if (!flag) break;
    }
    if (flag)
      cout << "Yes\n";
    else
      cout << "NO\n";
  }
  return 0;
}

### Prompt

Generate 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>
using namespace std;
int main() {
  string str;
  cin >> str;
  int n, flag, len;
  cin >> n;
  if (str.length() % n != 0 || str.length() < n)
    cout << "NO\n";
  else {
    len = str.length() / n;
    for (int i = 0; i < n; i++) {
      int last = (i + 1) * len - 1;
      flag = 1;
      for (int j = i * len; j < last; j++) {
        if (str[j] == str[last])
          last--;
        else {
          flag = 0;
          break;
        }
      }
      if (!flag) break;
    }
    if (flag)
      cout << "Yes\n";
    else
      cout << "NO\n";
  }
  return 0;
}

```