#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int n, a;
  while (cin >> s >> n) {
    a = s.length();
    if (a % n != 0)
      cout << "NO" << endl;
    else {
      int b, c = 0;
      b = a / n;
      for (int i = 0; i < a; i += b) {
        for (int j = 0; j < b / 2; j++) {
          if (s[i + j] != s[i + b - j - 1]) {
            c = 1;
            break;
          }
        }
        if (c == 1) break;
      }
      if (c == 1)
        cout << "NO" << endl;
      else
        cout << "YES" << endl;
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 s;
  int n, a;
  while (cin >> s >> n) {
    a = s.length();
    if (a % n != 0)
      cout << "NO" << endl;
    else {
      int b, c = 0;
      b = a / n;
      for (int i = 0; i < a; i += b) {
        for (int j = 0; j < b / 2; j++) {
          if (s[i + j] != s[i + b - j - 1]) {
            c = 1;
            break;
          }
        }
        if (c == 1) break;
      }
      if (c == 1)
        cout << "NO" << endl;
      else
        cout << "YES" << endl;
    }
  }
  return 0;
}

```