#include <bits/stdc++.h>
using namespace std;
const long long int maxn = 2 * 1e6 + 5;
const long long int modn = 1e9 + 7;
long long int Z, len, part;
string inp;
long long int all[maxn];
bool is_pal(long long int L, long long int R) {
  while (R > L) {
    if (inp[L] != inp[R]) return false;
    L++;
    R--;
  }
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> inp >> part;
  if (inp.size() % part != 0) {
    cout << "NO";
    return 0;
  }
  part = inp.size() / part;
  for (int i = 1; i <= inp.size(); i += part) {
    if (is_pal(i - 1, i + part - 2) == false) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES";
  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;
const long long int maxn = 2 * 1e6 + 5;
const long long int modn = 1e9 + 7;
long long int Z, len, part;
string inp;
long long int all[maxn];
bool is_pal(long long int L, long long int R) {
  while (R > L) {
    if (inp[L] != inp[R]) return false;
    L++;
    R--;
  }
  return true;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> inp >> part;
  if (inp.size() % part != 0) {
    cout << "NO";
    return 0;
  }
  part = inp.size() / part;
  for (int i = 1; i <= inp.size(); i += part) {
    if (is_pal(i - 1, i + part - 2) == false) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES";
  return 0;
}

```