#include <bits/stdc++.h>
using namespace std;
int main() {
  vector<string> mai;
  string input, s = "";
  int div;
  cin >> input >> div;
  int sz = input.size();
  int k = sz / div;
  if (sz % div != 0)
    cout << "NO" << endl;
  else {
    for (int i = 0; i < sz; i++) {
      k = sz / div;
      while (k) {
        s += input[i];
        k--;
        i++;
      }
      mai.push_back(s);
      s = "";
      i--;
    }
    for (int i = 0; i < mai.size(); i++) {
      if (mai[i] != string(mai[i].rbegin(), mai[i].rend())) {
        cout << "NO" << endl;
        return 0;
      }
    }
    cout << "YES" << endl;
  }
}

### Prompt

Your challenge is to write 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() {
  vector<string> mai;
  string input, s = "";
  int div;
  cin >> input >> div;
  int sz = input.size();
  int k = sz / div;
  if (sz % div != 0)
    cout << "NO" << endl;
  else {
    for (int i = 0; i < sz; i++) {
      k = sz / div;
      while (k) {
        s += input[i];
        k--;
        i++;
      }
      mai.push_back(s);
      s = "";
      i--;
    }
    for (int i = 0; i < mai.size(); i++) {
      if (mai[i] != string(mai[i].rbegin(), mai[i].rend())) {
        cout << "NO" << endl;
        return 0;
      }
    }
    cout << "YES" << endl;
  }
}

```