#include <bits/stdc++.h>
using namespace std;
int palindrom(const char* str) {
  int L = 0;
  for (L = 0; str[L] != '\0'; L++)
    ;
  int C = 0;
  for (C = 0; (C < L / 2) && (str[C] == str[L - C - 1]); C++)
    ;
  if (C == L / 2)
    return 1;
  else
    return 0;
}
int main() {
  char str[1000];
  char temp[1000];
  int k;
  cin >> str;
  cin >> k;
  int i = 0;
  int j = 0;
  int length = 0;
  for (length = 0; str[length] != '\0'; length++)
    ;
  if ((length % k) != 0) {
    cout << "NO";
  } else {
    while (i < length) {
      while (j < length / k) temp[j++] = str[i++];
      temp[j] = '\0';
      if (palindrom(temp) == 0) {
        cout << "NO";
        return 0;
      }
      j = 0;
    }
    cout << "YES";
  }
  return 0;
}

### Prompt

Create a solution in cpp for 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 palindrom(const char* str) {
  int L = 0;
  for (L = 0; str[L] != '\0'; L++)
    ;
  int C = 0;
  for (C = 0; (C < L / 2) && (str[C] == str[L - C - 1]); C++)
    ;
  if (C == L / 2)
    return 1;
  else
    return 0;
}
int main() {
  char str[1000];
  char temp[1000];
  int k;
  cin >> str;
  cin >> k;
  int i = 0;
  int j = 0;
  int length = 0;
  for (length = 0; str[length] != '\0'; length++)
    ;
  if ((length % k) != 0) {
    cout << "NO";
  } else {
    while (i < length) {
      while (j < length / k) temp[j++] = str[i++];
      temp[j] = '\0';
      if (palindrom(temp) == 0) {
        cout << "NO";
        return 0;
      }
      j = 0;
    }
    cout << "YES";
  }
  return 0;
}

```