#include <bits/stdc++.h>
using namespace std;
bool ev(string a, int len) {
  for (int i = 0; i < len; i++) {
    if (a[i] != a[len - i - 1]) {
      return false;
    }
  }
  return true;
}
int main() {
  string a, rpta = "YES";
  int k;
  cin >> a >> k;
  int len = a.size(), au = len / k;
  string v[k];
  if (len % k == 0) {
    for (int i = 0; i < len; i += au) {
      for (int j = i; j < i + au; j++) {
        v[i / au] += a[j];
      }
      if (!ev(v[i / au], au)) {
        rpta = "NO";
        break;
      }
    }
  } else {
    rpta = "NO";
  }
  cout << rpta << "\n";
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
bool ev(string a, int len) {
  for (int i = 0; i < len; i++) {
    if (a[i] != a[len - i - 1]) {
      return false;
    }
  }
  return true;
}
int main() {
  string a, rpta = "YES";
  int k;
  cin >> a >> k;
  int len = a.size(), au = len / k;
  string v[k];
  if (len % k == 0) {
    for (int i = 0; i < len; i += au) {
      for (int j = i; j < i + au; j++) {
        v[i / au] += a[j];
      }
      if (!ev(v[i / au], au)) {
        rpta = "NO";
        break;
      }
    }
  } else {
    rpta = "NO";
  }
  cout << rpta << "\n";
  return 0;
}

```