#include <bits/stdc++.h>
using namespace std;
int main() {
  string x, y, z;
  cin >> x;
  long f, s;
  long i, j;
  cin >> f;
  if (x.size() % f != 0) {
    goto h;
  }
  s = x.size() / f;
  for (i = 0; i < x.size(); i += s - 1) {
    for (j = i + s - 1; j >= i; j--) {
      y = y + x.substr(j, 1);
    }
    i++;
  }
  if (y == x) {
    cout << "YES";
  } else {
  h:
    cout << "NO";
  }
}

### 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;
int main() {
  string x, y, z;
  cin >> x;
  long f, s;
  long i, j;
  cin >> f;
  if (x.size() % f != 0) {
    goto h;
  }
  s = x.size() / f;
  for (i = 0; i < x.size(); i += s - 1) {
    for (j = i + s - 1; j >= i; j--) {
      y = y + x.substr(j, 1);
    }
    i++;
  }
  if (y == x) {
    cout << "YES";
  } else {
  h:
    cout << "NO";
  }
}

```