#include <bits/stdc++.h>
using namespace std;
string s;
int k, len;
bool isPalindrome(int a, int b) {
  for (int i = 0; i < (len); i++)
    if (s[a + i] != s[b - i]) return false;
  return true;
}
int main() {
  cin >> s;
  scanf("%d", &k);
  if (s.size() % k != 0) {
    printf("NO\n");
    return 0;
  }
  len = s.size() / k;
  int pos = 0;
  for (int i = 0; i < (k); i++) {
    if (!isPalindrome(pos, pos + len - 1)) {
      printf("NO\n");
      return 0;
    }
    pos += len;
  }
  printf("YES\n");
  return 0;
}

### Prompt

Your challenge is to write 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 k, len;
bool isPalindrome(int a, int b) {
  for (int i = 0; i < (len); i++)
    if (s[a + i] != s[b - i]) return false;
  return true;
}
int main() {
  cin >> s;
  scanf("%d", &k);
  if (s.size() % k != 0) {
    printf("NO\n");
    return 0;
  }
  len = s.size() / k;
  int pos = 0;
  for (int i = 0; i < (k); i++) {
    if (!isPalindrome(pos, pos + len - 1)) {
      printf("NO\n");
      return 0;
    }
    pos += len;
  }
  printf("YES\n");
  return 0;
}

```