#include <bits/stdc++.h>
using namespace std;
string str;
int length;
int length_of_palindrome;
bool flag = true;
int begini, endi;
int num;
int counter = 0;
int main() {
  cin >> str >> length_of_palindrome;
  length = str.length();
  if (length % length_of_palindrome == 0) {
    num = length_of_palindrome;
    begini = 0;
    endi = length / length_of_palindrome - 1;
    while (num) {
      flag = true;
      for (int i = begini; i <= (endi + begini) / 2; i++) {
        if (str[i] != str[endi + begini - i]) {
          flag = false;
          break;
        }
      }
      if (flag) {
        counter++;
      }
      begini = endi + 1;
      endi = begini + (length / length_of_palindrome) - 1;
      num--;
    }
    cout << ((counter == length_of_palindrome) ? "YES" : "NO");
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 str;
int length;
int length_of_palindrome;
bool flag = true;
int begini, endi;
int num;
int counter = 0;
int main() {
  cin >> str >> length_of_palindrome;
  length = str.length();
  if (length % length_of_palindrome == 0) {
    num = length_of_palindrome;
    begini = 0;
    endi = length / length_of_palindrome - 1;
    while (num) {
      flag = true;
      for (int i = begini; i <= (endi + begini) / 2; i++) {
        if (str[i] != str[endi + begini - i]) {
          flag = false;
          break;
        }
      }
      if (flag) {
        counter++;
      }
      begini = endi + 1;
      endi = begini + (length / length_of_palindrome) - 1;
      num--;
    }
    cout << ((counter == length_of_palindrome) ? "YES" : "NO");
  } else {
    cout << "NO" << endl;
  }
  return 0;
}

```