#include <bits/stdc++.h>
using namespace std;
bool isPal(string s) {
  for (int i = 0; i < s.size() / 2; i++) {
    if (s[i] != s[s.size() - i - 1]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  int k;
  cin >> k;
  bool ok = false;
  if (s.size() % k == 0) {
    string base = s.substr(0, s.size() / k);
    ok = isPal(base);
    for (int i = 1; i < k; i++) {
      string a = s.substr(s.size() / k * i, s.size() / k);
      ok = ok && isPal(a);
    }
  }
  cout << (ok ? "YES" : "NO") << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp 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 isPal(string s) {
  for (int i = 0; i < s.size() / 2; i++) {
    if (s[i] != s[s.size() - i - 1]) return false;
  }
  return true;
}
int main() {
  ios_base::sync_with_stdio(false);
  string s;
  cin >> s;
  int k;
  cin >> k;
  bool ok = false;
  if (s.size() % k == 0) {
    string base = s.substr(0, s.size() / k);
    ok = isPal(base);
    for (int i = 1; i < k; i++) {
      string a = s.substr(s.size() / k * i, s.size() / k);
      ok = ok && isPal(a);
    }
  }
  cout << (ok ? "YES" : "NO") << "\n";
  return 0;
}

```