#include <bits/stdc++.h>
using namespace std;
int palindrome(string s, int l, int r) {
  int i;
  for (i = l; i < r; i++) {
    if (s[i] != s[r - i - 1 + l]) return 0;
  }
  return 1;
}
int main() {
  int n, i;
  string s;
  cin >> s >> n;
  int c = 0;
  if (s.size() % n == 0) {
    for (i = 0; i < s.size(); i += (s.size() / n)) {
      if (palindrome(s, i, i + (s.size() / n))) {
        c++;
      }
    }
    if (c == n)
      cout << "YES";
    else
      cout << "NO";
  } else {
    cout << "NO";
  }
  return 0;
}

### 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 palindrome(string s, int l, int r) {
  int i;
  for (i = l; i < r; i++) {
    if (s[i] != s[r - i - 1 + l]) return 0;
  }
  return 1;
}
int main() {
  int n, i;
  string s;
  cin >> s >> n;
  int c = 0;
  if (s.size() % n == 0) {
    for (i = 0; i < s.size(); i += (s.size() / n)) {
      if (palindrome(s, i, i + (s.size() / n))) {
        c++;
      }
    }
    if (c == n)
      cout << "YES";
    else
      cout << "NO";
  } else {
    cout << "NO";
  }
  return 0;
}

```