#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  int l, j, ftemp, stemp;
  double temp;
  bool check = true;
  cin >> s >> l;
  temp = s.size() / double(l);
  l = s.size() / l;
  if (temp != l) check = false;
  for (int i = 0, j = l - 1; i < s.size() && check && j < s.size();
       i += l, j += l) {
    ftemp = i;
    stemp = j;
    while (ftemp < stemp) {
      if (s[ftemp] != s[stemp]) {
        check = false;
        break;
      }
      ftemp++;
      stemp--;
    }
    if (!check) break;
  }
  if (check)
    cout << "YES";
  else
    cout << "NO";
  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;
int main() {
  string s;
  int l, j, ftemp, stemp;
  double temp;
  bool check = true;
  cin >> s >> l;
  temp = s.size() / double(l);
  l = s.size() / l;
  if (temp != l) check = false;
  for (int i = 0, j = l - 1; i < s.size() && check && j < s.size();
       i += l, j += l) {
    ftemp = i;
    stemp = j;
    while (ftemp < stemp) {
      if (s[ftemp] != s[stemp]) {
        check = false;
        break;
      }
      ftemp++;
      stemp--;
    }
    if (!check) break;
  }
  if (check)
    cout << "YES";
  else
    cout << "NO";
  return 0;
}

```