#include <bits/stdc++.h>
using namespace std;
bool isPalindrome(string &x, int i, int j) {
  while (i < j) {
    if (x.at(i) != x.at(j)) return false;
    i++;
    j--;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  string x;
  int k, n;
  cin >> x >> k;
  if (x.size() % k == 0) {
    n = x.size() / k;
    bool yes = true;
    for (int i = 0; i < x.size(); i += n) {
      if (!(isPalindrome(x, i, i + n - 1))) {
        yes = false;
        break;
      }
    }
    if (yes)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  } else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Please formulate 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;
bool isPalindrome(string &x, int i, int j) {
  while (i < j) {
    if (x.at(i) != x.at(j)) return false;
    i++;
    j--;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(0);
  string x;
  int k, n;
  cin >> x >> k;
  if (x.size() % k == 0) {
    n = x.size() / k;
    bool yes = true;
    for (int i = 0; i < x.size(); i += n) {
      if (!(isPalindrome(x, i, i + n - 1))) {
        yes = false;
        break;
      }
    }
    if (yes)
      cout << "YES" << endl;
    else
      cout << "NO" << endl;
  } else
    cout << "NO" << endl;
  return 0;
}

```