#include <bits/stdc++.h>
using namespace std;
char s[1001];
bool cal(int i, int k) {
  for (int a = i, b = i + k - 1; a < i + k / 2; a++, b--)
    if (s[a] != s[b]) return false;
  return true;
}
int main() {
  int n, k, i;
  bool ok = true;
  scanf("%s\n%d", s, &k);
  n = strlen(s);
  if (n % k != 0)
    printf("NO");
  else {
    for (i = 0; i < n; i += n / k) ok &= cal(i, n / k);
    if (ok)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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[1001];
bool cal(int i, int k) {
  for (int a = i, b = i + k - 1; a < i + k / 2; a++, b--)
    if (s[a] != s[b]) return false;
  return true;
}
int main() {
  int n, k, i;
  bool ok = true;
  scanf("%s\n%d", s, &k);
  n = strlen(s);
  if (n % k != 0)
    printf("NO");
  else {
    for (i = 0; i < n; i += n / k) ok &= cal(i, n / k);
    if (ok)
      printf("YES");
    else
      printf("NO");
  }
  return 0;
}

```