#include <bits/stdc++.h>
using namespace std;
int arr[1000000], ans, brr[1000000], num, len, n, m, a, b, k, sum, cnt, keep, w,
    x, y, z, l, r;
bool check;
string S;
int j;
bool ispal(int i) {
  string temp = S.substr(i, j);
  string t = temp;
  reverse(temp.begin(), temp.end());
  if (t == temp) return 1;
  return 0;
}
int main() {
  cin >> S >> k;
  len = S.size();
  if (len % k == 0) {
    j = len / k;
    for (int i = 0; i < len; i += j) {
      if (!ispal(i)) {
        puts("NO");
        return 0;
      }
    }
    puts("YES");
    return 0;
  } else
    puts("NO");
  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;
int arr[1000000], ans, brr[1000000], num, len, n, m, a, b, k, sum, cnt, keep, w,
    x, y, z, l, r;
bool check;
string S;
int j;
bool ispal(int i) {
  string temp = S.substr(i, j);
  string t = temp;
  reverse(temp.begin(), temp.end());
  if (t == temp) return 1;
  return 0;
}
int main() {
  cin >> S >> k;
  len = S.size();
  if (len % k == 0) {
    j = len / k;
    for (int i = 0; i < len; i += j) {
      if (!ispal(i)) {
        puts("NO");
        return 0;
      }
    }
    puts("YES");
    return 0;
  } else
    puts("NO");
  return 0;
}

```