#include <bits/stdc++.h>
using namespace std;
int i, j;
bool flag = true;
int main() {
  string str11;
  cin >> str11;
  int n;
  scanf("%d", &n);
  if (str11.size() % n != 0) {
    printf("NO");
  } else {
    int length1 = str11.size();
    int x = str11.size() / n;
    i = 0;
    while (i < length1) {
      string str11part = str11.substr(i, x);
      int z = str11part.size();
      for (j = 0; j < z; j++) {
        if (str11part[j] != str11part[z - j - 1]) {
          flag = false;
        }
      }
      i = i + x;
    }
    if (flag == true) {
      printf("YES");
    } else {
      printf("NO");
    }
  }
  return 0;
}

### Prompt

Your task is to create 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 i, j;
bool flag = true;
int main() {
  string str11;
  cin >> str11;
  int n;
  scanf("%d", &n);
  if (str11.size() % n != 0) {
    printf("NO");
  } else {
    int length1 = str11.size();
    int x = str11.size() / n;
    i = 0;
    while (i < length1) {
      string str11part = str11.substr(i, x);
      int z = str11part.size();
      for (j = 0; j < z; j++) {
        if (str11part[j] != str11part[z - j - 1]) {
          flag = false;
        }
      }
      i = i + x;
    }
    if (flag == true) {
      printf("YES");
    } else {
      printf("NO");
    }
  }
  return 0;
}

```