#include <bits/stdc++.h>
using namespace std;
string str;
bool chk(string str) {
  if (str.size() <= 1) return true;
  bool flag = true;
  for (int i = 0; i < str.size() / 2; i++) {
    if (str[i] != str[str.size() - 1 - i]) flag = false;
  }
  return flag;
}
int main() {
  cin >> str;
  int n;
  cin >> n;
  int len = str.size();
  if ((len % n) != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int per = (len / n);
  string tmp = "";
  for (int i = 0; i < len; i++) {
    tmp.push_back(str[i]);
    if (tmp.size() == per) {
      if (chk(tmp) == false) {
        cout << "NO" << endl;
        return 0;
      }
      tmp.clear();
    }
  }
  cout << "YES" << endl;
  return 0;
}

### Prompt

Create a solution in CPP for 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 str;
bool chk(string str) {
  if (str.size() <= 1) return true;
  bool flag = true;
  for (int i = 0; i < str.size() / 2; i++) {
    if (str[i] != str[str.size() - 1 - i]) flag = false;
  }
  return flag;
}
int main() {
  cin >> str;
  int n;
  cin >> n;
  int len = str.size();
  if ((len % n) != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int per = (len / n);
  string tmp = "";
  for (int i = 0; i < len; i++) {
    tmp.push_back(str[i]);
    if (tmp.size() == per) {
      if (chk(tmp) == false) {
        cout << "NO" << endl;
        return 0;
      }
      tmp.clear();
    }
  }
  cout << "YES" << endl;
  return 0;
}

```