#include <bits/stdc++.h>
using namespace std;
int main() {
  string s;
  cin >> s;
  int t;
  cin >> t;
  int n;
  n = s.size();
  if (n % t == 0) {
    char a[1001];
    for (int i = 1; i <= n; i++) {
      a[i] = s[i - 1];
    }
    bool flag;
    flag = true;
    int j;
    int k;
    for (int i = 1; i < n; i = i + n / t) {
      j = i;
      k = i + n / t - 1;
      while (j < k) {
        if (a[j] != a[k]) {
          flag = false;
          break;
        }
        j++;
        k--;
      }
      if (flag == false) {
        break;
      }
    }
    if (flag == true) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  } else {
    cout << "NO";
  }
  return 0;
}

### Prompt

Generate 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;
  cin >> s;
  int t;
  cin >> t;
  int n;
  n = s.size();
  if (n % t == 0) {
    char a[1001];
    for (int i = 1; i <= n; i++) {
      a[i] = s[i - 1];
    }
    bool flag;
    flag = true;
    int j;
    int k;
    for (int i = 1; i < n; i = i + n / t) {
      j = i;
      k = i + n / t - 1;
      while (j < k) {
        if (a[j] != a[k]) {
          flag = false;
          break;
        }
        j++;
        k--;
      }
      if (flag == false) {
        break;
      }
    }
    if (flag == true) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  } else {
    cout << "NO";
  }
  return 0;
}

```