#include <bits/stdc++.h>
using namespace std;
int isPal(string str) {
  int ss = str.size() - 1;
  for (int k = 0; k < str.size() / 2; k++) {
    if (str[k] != str[ss]) return false;
    ss--;
  }
  return true;
}
int main() {
  string s;
  int a, cc = 0;
  cin >> s;
  cin >> a;
  if (s.size() % a == 0) {
    int mov = s.size() / a;
    for (int k = 0; k < s.size(); k += mov) {
      if (!isPal(s.substr(k, mov))) {
        cout << "NO" << endl;
        return 0;
      }
    }
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}

### 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 isPal(string str) {
  int ss = str.size() - 1;
  for (int k = 0; k < str.size() / 2; k++) {
    if (str[k] != str[ss]) return false;
    ss--;
  }
  return true;
}
int main() {
  string s;
  int a, cc = 0;
  cin >> s;
  cin >> a;
  if (s.size() % a == 0) {
    int mov = s.size() / a;
    for (int k = 0; k < s.size(); k += mov) {
      if (!isPal(s.substr(k, mov))) {
        cout << "NO" << endl;
        return 0;
      }
    }
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
  }
}

```