#include <bits/stdc++.h>
using namespace std;
bool go(string s) {
  string p = s;
  reverse(p.begin(), p.end());
  return (s == p);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, i, j, n, m, l, r;
  string s;
  cin >> s >> l;
  n = s.size();
  if (n % l != 0)
    cout << "NO\n";
  else {
    l = n / l;
    int f = 0;
    for (i = 0; i < n && f == 0; i += l) {
      string p;
      for (j = i; j < i + l; j++) p += s[j];
      if (go(p) == false) f = 1;
    }
    cout << ((f == 0) ? "YES\n" : "NO\n");
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 go(string s) {
  string p = s;
  reverse(p.begin(), p.end());
  return (s == p);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, i, j, n, m, l, r;
  string s;
  cin >> s >> l;
  n = s.size();
  if (n % l != 0)
    cout << "NO\n";
  else {
    l = n / l;
    int f = 0;
    for (i = 0; i < n && f == 0; i += l) {
      string p;
      for (j = i; j < i + l; j++) p += s[j];
      if (go(p) == false) f = 1;
    }
    cout << ((f == 0) ? "YES\n" : "NO\n");
  }
  return 0;
}

```