#include <bits/stdc++.h>
using namespace std;
bool is_palindrome(const string& s) {
  int first = 0;
  int last = s.length() - 1;
  while (first < last) {
    if (s[first] != s[last]) return false;
    ++first;
    --last;
  }
  return true;
}
void A() {
  string str;
  int k, len;
  cin >> str >> k;
  if (str.length() % k) {
    cout << "NO" << endl;
    return;
  }
  len = str.length() / k;
  for (int i = 0; i < k; ++i) {
    if (!is_palindrome(str.substr(i * len, len))) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
}
int main() {
  A();
  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;
bool is_palindrome(const string& s) {
  int first = 0;
  int last = s.length() - 1;
  while (first < last) {
    if (s[first] != s[last]) return false;
    ++first;
    --last;
  }
  return true;
}
void A() {
  string str;
  int k, len;
  cin >> str >> k;
  if (str.length() % k) {
    cout << "NO" << endl;
    return;
  }
  len = str.length() / k;
  for (int i = 0; i < k; ++i) {
    if (!is_palindrome(str.substr(i * len, len))) {
      cout << "NO" << endl;
      return;
    }
  }
  cout << "YES" << endl;
}
int main() {
  A();
  return 0;
}

```