#include <bits/stdc++.h>
using namespace std;
int main(int argc, char** argv) {
  string s;
  int k;
  cin >> s >> k;
  int res = s.size() / k;
  bool check = true;
  bool palindrom = true;
  if (s.size() % k != 0)
    palindrom = false;
  else {
    for (int i = 0; i < k; i++) {
      check = true;
      palindrom = true;
      for (int j = 0; j < res / 2; j++) {
        if (s[j] == s[res - j - 1]) {
          palindrom = true;
        } else {
          palindrom = false;
          break;
        }
      }
      s.erase(s.begin(), s.begin() + res);
      if (palindrom == false) break;
    }
  }
  if (palindrom)
    cout << "YES" << 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;
int main(int argc, char** argv) {
  string s;
  int k;
  cin >> s >> k;
  int res = s.size() / k;
  bool check = true;
  bool palindrom = true;
  if (s.size() % k != 0)
    palindrom = false;
  else {
    for (int i = 0; i < k; i++) {
      check = true;
      palindrom = true;
      for (int j = 0; j < res / 2; j++) {
        if (s[j] == s[res - j - 1]) {
          palindrom = true;
        } else {
          palindrom = false;
          break;
        }
      }
      s.erase(s.begin(), s.begin() + res);
      if (palindrom == false) break;
    }
  }
  if (palindrom)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
  return 0;
}

```