#include <bits/stdc++.h>
using namespace std;
int main(void) {
  string s;
  unsigned int k;
  cin >> s;
  cin >> k;
  unsigned int nn = s.size() / k;
  bool pal = true;
  if (s.size() % k == 0) {
    for (auto i = 0u; i < k; i++) {
      for (auto j = 0u; j < nn; j++) {
        if (s.at(i * nn + j) != s.at(i * nn + (nn - j - 1))) {
          pal = false;
        }
      }
    }
  } else {
    pal = false;
  }
  if (pal) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  cout << endl;
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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(void) {
  string s;
  unsigned int k;
  cin >> s;
  cin >> k;
  unsigned int nn = s.size() / k;
  bool pal = true;
  if (s.size() % k == 0) {
    for (auto i = 0u; i < k; i++) {
      for (auto j = 0u; j < nn; j++) {
        if (s.at(i * nn + j) != s.at(i * nn + (nn - j - 1))) {
          pal = false;
        }
      }
    }
  } else {
    pal = false;
  }
  if (pal) {
    cout << "YES";
  } else {
    cout << "NO";
  }
  cout << endl;
  return 0;
}

```