#include <bits/stdc++.h>
using namespace std;
int main() {
  std::ios::sync_with_stdio(false);
  string str;
  cin >> str;
  int k, len, hlen, size;
  cin >> k;
  bool flag = true;
  len = (str.length()) / k;
  hlen = len / 2;
  if (k * len == str.length()) {
    for (__typeof(0) i = (0); i < (k); i++) {
      for (__typeof(0) j = (0); j < (hlen); j++) {
        if (str[i * len + j] != str[i * len + len - 1 - j]) {
          flag = false;
          break;
        }
      }
      if (!flag) break;
    }
    if (!flag)
      cout << "NO" << endl;
    else
      cout << "YES" << endl;
  } else
    cout << "NO" << endl;
  return 0;
}

### Prompt

Create a solution in Cpp for 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() {
  std::ios::sync_with_stdio(false);
  string str;
  cin >> str;
  int k, len, hlen, size;
  cin >> k;
  bool flag = true;
  len = (str.length()) / k;
  hlen = len / 2;
  if (k * len == str.length()) {
    for (__typeof(0) i = (0); i < (k); i++) {
      for (__typeof(0) j = (0); j < (hlen); j++) {
        if (str[i * len + j] != str[i * len + len - 1 - j]) {
          flag = false;
          break;
        }
      }
      if (!flag) break;
    }
    if (!flag)
      cout << "NO" << endl;
    else
      cout << "YES" << endl;
  } else
    cout << "NO" << endl;
  return 0;
}

```