#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int k;
  cin >> s;
  cin >> k;
  if (s.length() % k != 0) {
    cout << "NO\n";
    return 0;
  }
  int n = s.length() / k;
  bool con = 0;
  for (int i = int(0); i <= int(k - 1); i++) {
    int l = i * n;
    int r = l + n - 1;
    while (l < r) {
      if (s[l] != s[r]) {
        con = 1;
        break;
      }
      l++;
      r--;
    }
    if (con) break;
  }
  if (!con)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

### 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;
int main() {
  string s;
  int k;
  cin >> s;
  cin >> k;
  if (s.length() % k != 0) {
    cout << "NO\n";
    return 0;
  }
  int n = s.length() / k;
  bool con = 0;
  for (int i = int(0); i <= int(k - 1); i++) {
    int l = i * n;
    int r = l + n - 1;
    while (l < r) {
      if (s[l] != s[r]) {
        con = 1;
        break;
      }
      l++;
      r--;
    }
    if (con) break;
  }
  if (!con)
    cout << "YES\n";
  else
    cout << "NO\n";
  return 0;
}

```