#include <bits/stdc++.h>
using namespace std;
bool pal(string s, int start, int end) {
  if (s[start] != s[end])
    return 0;
  else if (start >= end)
    return 1;
  else
    return pal(s, ++start, --end);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string merged;
  cin >> merged;
  int parts;
  cin >> parts;
  int sz = merged.size();
  if (sz % parts != 0) {
    cout << "NO" << endl;
  } else {
    int i = 0, loc = 0;
    int div = sz / parts;
    while (i < parts) {
      if (pal(merged.substr(loc, div), 0, div - 1)) {
        loc += div;
      } else {
        cout << "NO" << endl;
        return 0;
      }
      i++;
    }
    cout << "YES" << endl;
  }
  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 pal(string s, int start, int end) {
  if (s[start] != s[end])
    return 0;
  else if (start >= end)
    return 1;
  else
    return pal(s, ++start, --end);
  return 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string merged;
  cin >> merged;
  int parts;
  cin >> parts;
  int sz = merged.size();
  if (sz % parts != 0) {
    cout << "NO" << endl;
  } else {
    int i = 0, loc = 0;
    int div = sz / parts;
    while (i < parts) {
      if (pal(merged.substr(loc, div), 0, div - 1)) {
        loc += div;
      } else {
        cout << "NO" << endl;
        return 0;
      }
      i++;
    }
    cout << "YES" << endl;
  }
  return 0;
}

```