#include <bits/stdc++.h>
using namespace std;
int main() {
  char in[1000];
  int n, flag = 0, k = 0, r = 0, val = 0, u, v;
  gets(in);
  cin >> n;
  int count = n;
  int x = strlen(in);
  k = 0;
  r = x / n - 1;
  if (x == 1 && n <= x) {
    flag = 1;
    count = 0;
    x = 0;
  }
  while (r < x && flag == 0) {
    if (in[k] == in[r]) {
      for (u = k, v = r; u < v; u++, v--) {
        if (in[u] != in[v]) {
          val = 1;
          break;
        }
      }
      --count;
      k += x / n;
      r += x / n;
    } else {
      val = 1;
      flag = 1;
    }
  }
  if (val == 0 && count == 0 && k >= x)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

### Prompt

Your task is to create 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;
int main() {
  char in[1000];
  int n, flag = 0, k = 0, r = 0, val = 0, u, v;
  gets(in);
  cin >> n;
  int count = n;
  int x = strlen(in);
  k = 0;
  r = x / n - 1;
  if (x == 1 && n <= x) {
    flag = 1;
    count = 0;
    x = 0;
  }
  while (r < x && flag == 0) {
    if (in[k] == in[r]) {
      for (u = k, v = r; u < v; u++, v--) {
        if (in[u] != in[v]) {
          val = 1;
          break;
        }
      }
      --count;
      k += x / n;
      r += x / n;
    } else {
      val = 1;
      flag = 1;
    }
  }
  if (val == 0 && count == 0 && k >= x)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```