#include <bits/stdc++.h>
const int maxBufSize = (10000000);
using namespace std;
long long int k;
string s;
int main() {
  long long int i, j, a, b, c, len, slen;
  cin >> s >> k;
  len = s.length();
  if (len % k == 0) {
    c = 0;
    slen = len / k;
    for (i = 1; i <= k; ++i) {
      a = (i - 1) * slen;
      b = a + slen;
      for (j = a; j <= (b + a) / 2 - 1; ++j) {
        if (s[j] != s[a + b - j - 1]) {
          c = 1;
          break;
        }
      }
      if (c == 1) {
        break;
      }
    }
    if (c == 0) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  } else {
    cout << "NO";
  }
  return 0;
}

### Prompt

In CPP, your task is to solve 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>
const int maxBufSize = (10000000);
using namespace std;
long long int k;
string s;
int main() {
  long long int i, j, a, b, c, len, slen;
  cin >> s >> k;
  len = s.length();
  if (len % k == 0) {
    c = 0;
    slen = len / k;
    for (i = 1; i <= k; ++i) {
      a = (i - 1) * slen;
      b = a + slen;
      for (j = a; j <= (b + a) / 2 - 1; ++j) {
        if (s[j] != s[a + b - j - 1]) {
          c = 1;
          break;
        }
      }
      if (c == 1) {
        break;
      }
    }
    if (c == 0) {
      cout << "YES";
    } else {
      cout << "NO";
    }
  } else {
    cout << "NO";
  }
  return 0;
}

```