#include <bits/stdc++.h>
using namespace std;
char s[1010];
int n, i, q, k, l, r;
bool ok;
int main() {
  scanf("%s%d", s, &k);
  n = strlen(s);
  if (n % k) {
    printf("NO\n");
    return 0;
  }
  q = n / k;
  ok = true;
  for (i = q; i <= n; i += q) {
    for (l = i - q, r = i - 1; l <= r; l++, r--) {
      if (s[r] == s[l]) continue;
      ok = false;
    }
    if (ok == false) break;
  }
  if (ok == true) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}

### Prompt

Generate 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;
char s[1010];
int n, i, q, k, l, r;
bool ok;
int main() {
  scanf("%s%d", s, &k);
  n = strlen(s);
  if (n % k) {
    printf("NO\n");
    return 0;
  }
  q = n / k;
  ok = true;
  for (i = q; i <= n; i += q) {
    for (l = i - q, r = i - 1; l <= r; l++, r--) {
      if (s[r] == s[l]) continue;
      ok = false;
    }
    if (ok == false) break;
  }
  if (ok == true) {
    printf("YES\n");
    return 0;
  }
  printf("NO\n");
  return 0;
}

```