#include <bits/stdc++.h>
using namespace std;
bool is_palin(string s, int i, int j, int size) {
  for (int k = i, l = j; k <= l; k++, l--) {
    if (s[k] != s[l]) return false;
  }
  return true;
}
int main() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int size = s.size();
  if (size % k != 0) {
    cout << "NO" << endl;
  } else {
    int ini = 0;
    int fin = (size / k) - 1;
    bool c = true;
    for (int i = 0; i < k; i++) {
      if (!is_palin(s, ini, fin, size)) {
        cout << "NO" << endl;
        c = false;
        break;
      }
      int temp = fin;
      ini = fin + 1;
      fin = temp + (size) / k;
    }
    if (c) {
      cout << "YES" << endl;
    }
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 is_palin(string s, int i, int j, int size) {
  for (int k = i, l = j; k <= l; k++, l--) {
    if (s[k] != s[l]) return false;
  }
  return true;
}
int main() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int size = s.size();
  if (size % k != 0) {
    cout << "NO" << endl;
  } else {
    int ini = 0;
    int fin = (size / k) - 1;
    bool c = true;
    for (int i = 0; i < k; i++) {
      if (!is_palin(s, ini, fin, size)) {
        cout << "NO" << endl;
        c = false;
        break;
      }
      int temp = fin;
      ini = fin + 1;
      fin = temp + (size) / k;
    }
    if (c) {
      cout << "YES" << endl;
    }
  }
  return 0;
}

```