#include <bits/stdc++.h>
using namespace std;
string s, a[1007];
int n, l, t, r;
bool tt = true;
int palindrom(string d) {
  for (int i = 0; i < t / 2; i++) {
    if (d[i] != d[t - (i + 1)]) {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> s >> n;
  l = s.length();
  if (l % n != 0)
    cout << "NO";
  else {
    t = l / n;
    for (int i = 0; i < l; i++) {
      if ((i + 1) % t == 0)
        a[r] += s[i], r++;
      else
        a[r] += s[i];
    }
    for (int i = 0; i < r; i++) {
      if (palindrom(a[i]) == 0) {
        cout << "NO";
        return 0;
      }
    }
    cout << "YES";
  }
  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, a[1007];
int n, l, t, r;
bool tt = true;
int palindrom(string d) {
  for (int i = 0; i < t / 2; i++) {
    if (d[i] != d[t - (i + 1)]) {
      return 0;
    }
  }
  return 1;
}
int main() {
  cin >> s >> n;
  l = s.length();
  if (l % n != 0)
    cout << "NO";
  else {
    t = l / n;
    for (int i = 0; i < l; i++) {
      if ((i + 1) % t == 0)
        a[r] += s[i], r++;
      else
        a[r] += s[i];
    }
    for (int i = 0; i < r; i++) {
      if (palindrom(a[i]) == 0) {
        cout << "NO";
        return 0;
      }
    }
    cout << "YES";
  }
  return 0;
}

```