#include <bits/stdc++.h>
using namespace std;
int main() {
  int k, i, j;
  string a, b = "", c = "";
  bool u = true;
  cin >> a >> k;
  if (a.size() % k != 0)
    cout << "NO";
  else {
    int l = a.size() / k;
    for (i = 0; i < a.size(); i += l) {
      b = "";
      c = "";
      for (j = i; j < l + i; j++) {
        b += a[j];
        c = a[j] + c;
      }
      if (b != c) {
        u = false;
        break;
      }
    }
    if (u)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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() {
  int k, i, j;
  string a, b = "", c = "";
  bool u = true;
  cin >> a >> k;
  if (a.size() % k != 0)
    cout << "NO";
  else {
    int l = a.size() / k;
    for (i = 0; i < a.size(); i += l) {
      b = "";
      c = "";
      for (j = i; j < l + i; j++) {
        b += a[j];
        c = a[j] + c;
      }
      if (b != c) {
        u = false;
        break;
      }
    }
    if (u)
      cout << "YES";
    else
      cout << "NO";
  }
  return 0;
}

```