#include <bits/stdc++.h>
using namespace std;
string second;
long long int k;
long long int len;
long long int l, r;
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> (second) >> (k);
  len = second.size() / k;
  if (second.size() % k != 0) {
    cout << ("NO") << endl;
    return 0;
  }
  for (long long int i = 0; i < k; i++) {
    l = i * len;
    r = (i + 1) * len - 1;
    long long int iniL = l;
    for (; r >= iniL; l++, r--) {
      if (second[l] != second[r]) {
        cout << ("NO") << endl;
        return 0;
      }
    }
  }
  cout << ("YES") << endl;
  return 0;
}

### Prompt

Please formulate 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 second;
long long int k;
long long int len;
long long int l, r;
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> (second) >> (k);
  len = second.size() / k;
  if (second.size() % k != 0) {
    cout << ("NO") << endl;
    return 0;
  }
  for (long long int i = 0; i < k; i++) {
    l = i * len;
    r = (i + 1) * len - 1;
    long long int iniL = l;
    for (; r >= iniL; l++, r--) {
      if (second[l] != second[r]) {
        cout << ("NO") << endl;
        return 0;
      }
    }
  }
  cout << ("YES") << endl;
  return 0;
}

```