#include <bits/stdc++.h>
using namespace std;
bool chk(string s) {
  int n = s.length();
  for (int i = 0; i < n / 2; i++)
    if (s[i] != s[n - i - 1]) return false;
  return true;
}
int main() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int n = s.length();
  if (n % k != 0) {
    cout << "NO\n";
    return 0;
  }
  bool flag = true;
  int j = 0;
  for (int i = 0; i < k; i++) {
    string A = "";
    for (int q = 0; q < n / k; q++) {
      A.push_back(s[q + j]);
    }
    if (!chk(A)) flag = false;
    j = j + n / k;
  }
  if (flag) {
    cout << "YES\n";
  } else
    cout << "NO\n";
  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;
bool chk(string s) {
  int n = s.length();
  for (int i = 0; i < n / 2; i++)
    if (s[i] != s[n - i - 1]) return false;
  return true;
}
int main() {
  string s;
  cin >> s;
  int k;
  cin >> k;
  int n = s.length();
  if (n % k != 0) {
    cout << "NO\n";
    return 0;
  }
  bool flag = true;
  int j = 0;
  for (int i = 0; i < k; i++) {
    string A = "";
    for (int q = 0; q < n / k; q++) {
      A.push_back(s[q + j]);
    }
    if (!chk(A)) flag = false;
    j = j + n / k;
  }
  if (flag) {
    cout << "YES\n";
  } else
    cout << "NO\n";
  return 0;
}

```