#include <bits/stdc++.h>
using namespace std;
bool isPalin(string abcd) {
  int i = 0, j = abcd.size() - 1;
  while (i < j) {
    if (abcd[i] != abcd[j]) return false;
    i++;
    j--;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int k, sz = 0;
  string input;
  cin >> input >> k;
  if (input.size() % k != 0)
    cout << "NO\n";
  else {
    sz = input.size() / k;
    for (int i = 0; i < input.size(); i += sz) {
      if (!isPalin(input.substr(i, sz))) {
        cout << "NO\n";
        exit(0);
      }
    }
    cout << "YES\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;
bool isPalin(string abcd) {
  int i = 0, j = abcd.size() - 1;
  while (i < j) {
    if (abcd[i] != abcd[j]) return false;
    i++;
    j--;
  }
  return true;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int k, sz = 0;
  string input;
  cin >> input >> k;
  if (input.size() % k != 0)
    cout << "NO\n";
  else {
    sz = input.size() / k;
    for (int i = 0; i < input.size(); i += sz) {
      if (!isPalin(input.substr(i, sz))) {
        cout << "NO\n";
        exit(0);
      }
    }
    cout << "YES\n";
  }
  return 0;
}

```