#include <bits/stdc++.h>
using namespace std;
bool ispalindrome(string s) {
  long long int i = 0, j = s.size() - 1;
  while (i <= j) {
    if (s[i] != s[j]) return false;
    i++;
    j--;
  }
  return true;
}
void solve() {
  string s;
  cin >> s;
  long long int k;
  cin >> k;
  long long int n = s.size();
  if (n % k != 0) {
    cout << "NO" << endl;
    return;
  }
  for (long long int i = 0; i < n; i += (n / k)) {
    string p = s.substr(i, (n / k));
    if (ispalindrome(p) == false) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
  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;
bool ispalindrome(string s) {
  long long int i = 0, j = s.size() - 1;
  while (i <= j) {
    if (s[i] != s[j]) return false;
    i++;
    j--;
  }
  return true;
}
void solve() {
  string s;
  cin >> s;
  long long int k;
  cin >> k;
  long long int n = s.size();
  if (n % k != 0) {
    cout << "NO" << endl;
    return;
  }
  for (long long int i = 0; i < n; i += (n / k)) {
    string p = s.substr(i, (n / k));
    if (ispalindrome(p) == false) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long int t = 1;
  while (t--) {
    solve();
  }
  return 0;
}

```