#include <bits/stdc++.h>
using namespace std;
bool palindrome(string s, long long i, long long j) {
  long long I, coun = 0, t;
  t = i + (j - i + 1) / 2;
  for (I = i; I < t; I++) {
    if (s[I] != s[j]) {
      coun++;
      break;
    }
    j--;
  }
  if (coun == 0)
    return true;
  else
    return false;
}
int main() {
  long long k, coun = 0;
  string s;
  cin >> s;
  cin >> k;
  long long i, j, I, l;
  I = 0;
  j = s.length() / k - 1;
  if (s.length() % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  l = j - I + 1;
  for (i = 0; i < k; i++) {
    if (palindrome(s, I, j) == 0) {
      coun++;
      break;
    }
    I += l;
    j += l;
  }
  if (coun == 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

### Prompt

Please formulate 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 palindrome(string s, long long i, long long j) {
  long long I, coun = 0, t;
  t = i + (j - i + 1) / 2;
  for (I = i; I < t; I++) {
    if (s[I] != s[j]) {
      coun++;
      break;
    }
    j--;
  }
  if (coun == 0)
    return true;
  else
    return false;
}
int main() {
  long long k, coun = 0;
  string s;
  cin >> s;
  cin >> k;
  long long i, j, I, l;
  I = 0;
  j = s.length() / k - 1;
  if (s.length() % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  l = j - I + 1;
  for (i = 0; i < k; i++) {
    if (palindrome(s, I, j) == 0) {
      coun++;
      break;
    }
    I += l;
    j += l;
  }
  if (coun == 0)
    cout << "YES" << endl;
  else
    cout << "NO" << endl;
}

```