#include <bits/stdc++.h>
using namespace std;
int check_palin(string s) {
  int l = s.length();
  int flag = 0;
  for (int i = 0; i < l / 2; i++) {
    if (s[i] != s[l - i - 1]) {
      flag = 1;
      break;
    }
  }
  if (flag == 0)
    return (1);
  else
    return (0);
}
int main() {
  string s;
  cin >> s;
  int k, p;
  cin >> k;
  int j = 0, flag = 0;
  string sub;
  int l = s.length();
  if (l % k != 0)
    cout << "NO" << endl;
  else {
    for (int i = 0; i < k; i++) {
      p = 0;
      sub = s.substr(j, l / k);
      j += l / k;
      p = check_palin(sub);
      if (p == 0) {
        cout << "NO" << endl;
        flag = 1;
        break;
      }
    }
    if (flag == 0) cout << "YES" << endl;
  }
}

### 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;
int check_palin(string s) {
  int l = s.length();
  int flag = 0;
  for (int i = 0; i < l / 2; i++) {
    if (s[i] != s[l - i - 1]) {
      flag = 1;
      break;
    }
  }
  if (flag == 0)
    return (1);
  else
    return (0);
}
int main() {
  string s;
  cin >> s;
  int k, p;
  cin >> k;
  int j = 0, flag = 0;
  string sub;
  int l = s.length();
  if (l % k != 0)
    cout << "NO" << endl;
  else {
    for (int i = 0; i < k; i++) {
      p = 0;
      sub = s.substr(j, l / k);
      j += l / k;
      p = check_palin(sub);
      if (p == 0) {
        cout << "NO" << endl;
        flag = 1;
        break;
      }
    }
    if (flag == 0) cout << "YES" << endl;
  }
}

```