#include <bits/stdc++.h>
using namespace std;
string mes[1005];
bool check(string message) {
  string s;
  int len_check = message.length();
  len_check--;
  for (int i = 0; i <= len_check; i++) {
    if (message[i] != message[len_check - i]) return false;
  }
  return true;
}
int main() {
  bool YES = true;
  string s;
  int k;
  cin >> s >> k;
  int len = s.length();
  int per_len = len / k;
  if (len % k != 0) YES = false;
  if (YES)
    for (int i = 0; i < k; i++) {
      mes[i] = s.substr(i * per_len, per_len);
      if (!check(mes[i])) YES = false;
    }
  if (YES)
    puts("YES");
  else
    puts("NO");
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 mes[1005];
bool check(string message) {
  string s;
  int len_check = message.length();
  len_check--;
  for (int i = 0; i <= len_check; i++) {
    if (message[i] != message[len_check - i]) return false;
  }
  return true;
}
int main() {
  bool YES = true;
  string s;
  int k;
  cin >> s >> k;
  int len = s.length();
  int per_len = len / k;
  if (len % k != 0) YES = false;
  if (YES)
    for (int i = 0; i < k; i++) {
      mes[i] = s.substr(i * per_len, per_len);
      if (!check(mes[i])) YES = false;
    }
  if (YES)
    puts("YES");
  else
    puts("NO");
  return 0;
}

```