#include <bits/stdc++.h>
using namespace std;
int brr[100];
int main() {
  string s;
  cin >> s;
  int n, cnt = 0, cnt1 = 0, v;
  cin >> n;
  int len = s.length();
  int num = len / n;
  int start = 0;
  int end = num - 1;
  if (len % n != 0) {
    cout << "NO";
  } else {
    while (n) {
      v = 0;
      while (start < end) {
        if (s[start] == s[end]) {
          start++;
          end--;
          v++;
          continue;
        } else {
          cnt = 1;
          break;
        }
      }
      if (cnt == 1) {
        cout << "NO";
        cnt1++;
        break;
      }
      start -= v;
      end += v;
      start += num;
      end += num;
      n--;
    }
    if (cnt1 == 0) cout << "YES";
  }
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 brr[100];
int main() {
  string s;
  cin >> s;
  int n, cnt = 0, cnt1 = 0, v;
  cin >> n;
  int len = s.length();
  int num = len / n;
  int start = 0;
  int end = num - 1;
  if (len % n != 0) {
    cout << "NO";
  } else {
    while (n) {
      v = 0;
      while (start < end) {
        if (s[start] == s[end]) {
          start++;
          end--;
          v++;
          continue;
        } else {
          cnt = 1;
          break;
        }
      }
      if (cnt == 1) {
        cout << "NO";
        cnt1++;
        break;
      }
      start -= v;
      end += v;
      start += num;
      end += num;
      n--;
    }
    if (cnt1 == 0) cout << "YES";
  }
  return 0;
}

```