#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
bool f(string s) {
  string t = s;
  reverse(t.begin(), t.end());
  return (int)t.size() == (int)s.size() && t == s;
}
int main() {
  string s, t;
  int n, k;
  cin >> s >> k;
  s = '#' + s;
  n = (int)s.size() - 1;
  if (n % k) return cout << "NO\n", 0;
  for (int i = 1; i <= n; ++i) {
    t += s[i];
    if ((int)t.size() == n / k) {
      if (!f(t)) return cout << "NO", 0;
      t = "";
    }
  }
  cout << "YES";
}

### Prompt

Your task is to create 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;
const int N = 1e5 + 5;
bool f(string s) {
  string t = s;
  reverse(t.begin(), t.end());
  return (int)t.size() == (int)s.size() && t == s;
}
int main() {
  string s, t;
  int n, k;
  cin >> s >> k;
  s = '#' + s;
  n = (int)s.size() - 1;
  if (n % k) return cout << "NO\n", 0;
  for (int i = 1; i <= n; ++i) {
    t += s[i];
    if ((int)t.size() == n / k) {
      if (!f(t)) return cout << "NO", 0;
      t = "";
    }
  }
  cout << "YES";
}

```