#include <bits/stdc++.h>
using namespace std;
string s;
int check_palin(int x, int y) {
  for (int i = 0; i <= (y - x) / 2; i++) {
    if (s[x + i] != s[y - i]) return 0;
  }
  return 1;
}
int main(void) {
  int k, ans = 1;
  cin >> s >> k;
  int x = s.length() / k;
  if (s.length() % k) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 0; i < s.length(); i += x) {
    cerr << i << " " << i + x - 1 << " " << endl;
    if (!check_palin(i, i + x - 1)) ans = 0;
  }
  cout << (ans ? "YES" : "NO");
}

### 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;
string s;
int check_palin(int x, int y) {
  for (int i = 0; i <= (y - x) / 2; i++) {
    if (s[x + i] != s[y - i]) return 0;
  }
  return 1;
}
int main(void) {
  int k, ans = 1;
  cin >> s >> k;
  int x = s.length() / k;
  if (s.length() % k) {
    cout << "NO\n";
    return 0;
  }
  for (int i = 0; i < s.length(); i += x) {
    cerr << i << " " << i + x - 1 << " " << endl;
    if (!check_palin(i, i + x - 1)) ans = 0;
  }
  cout << (ans ? "YES" : "NO");
}

```