#include <bits/stdc++.h>
using namespace std;
int main(int argc, char const *argv[]) {
  string s;
  int n, len, i;
  cin >> s;
  cin >> n;
  len = s.size();
  if (len % n) {
    cout << "NO" << endl;
  } else {
    int each = len / n;
    int j = 0;
    string temp, temp1;
    int fl = 0;
    for (i = 0; i < n; i++) {
      temp = s.substr(j, each);
      temp1 = string(temp.rbegin(), temp.rend());
      if (temp != temp1) {
        fl = 1;
        break;
      }
      j = j + each;
    }
    if (fl) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
    }
  }
  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;
int main(int argc, char const *argv[]) {
  string s;
  int n, len, i;
  cin >> s;
  cin >> n;
  len = s.size();
  if (len % n) {
    cout << "NO" << endl;
  } else {
    int each = len / n;
    int j = 0;
    string temp, temp1;
    int fl = 0;
    for (i = 0; i < n; i++) {
      temp = s.substr(j, each);
      temp1 = string(temp.rbegin(), temp.rend());
      if (temp != temp1) {
        fl = 1;
        break;
      }
      j = j + each;
    }
    if (fl) {
      cout << "NO" << endl;
    } else {
      cout << "YES" << endl;
    }
  }
  return 0;
}

```