#include <bits/stdc++.h>
using namespace std;
int main() {
  string x, y;
  int n, z, i = 0, a = 0;
  cin >> x >> n;
  if ((int)x.size() % n != 0) {
    cout << "NO";
    return 0;
  }
  z = x.size() / n;
  while (i < x.size()) {
    y = x.substr(i, x.size() / n);
    if (y == string(y.rbegin(), y.rend())) {
      a++;
      i += z;
    } else
      break;
  }
  if (a == n)
    cout << "YES" << endl;
  else
    cout << "NO";
}

### 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 x, y;
  int n, z, i = 0, a = 0;
  cin >> x >> n;
  if ((int)x.size() % n != 0) {
    cout << "NO";
    return 0;
  }
  z = x.size() / n;
  while (i < x.size()) {
    y = x.substr(i, x.size() / n);
    if (y == string(y.rbegin(), y.rend())) {
      a++;
      i += z;
    } else
      break;
  }
  if (a == n)
    cout << "YES" << endl;
  else
    cout << "NO";
}

```