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

```