#include <bits/stdc++.h>
using namespace std;
static const int MAXN = 1e3 + 10;
int k;
char data[MAXN];
int main(int argc, char const *argv[]) {
  scanf(" %s", data + 1);
  int len = strlen(data + 1);
  scanf("%d", &k);
  if (len % k || k > len) {
    puts("NO");
    return 0;
  }
  int cut = len / k;
  bool flag = 1;
  for (int i = 1; i <= len; i += cut) {
    int l = i, r = i + cut - 1;
    while (l < r) {
      if (data[l] != data[r]) {
        flag = 0;
        break;
      }
      ++l, --r;
    }
    if (!flag) break;
  }
  if (!flag)
    puts("NO");
  else
    puts("YES");
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
static const int MAXN = 1e3 + 10;
int k;
char data[MAXN];
int main(int argc, char const *argv[]) {
  scanf(" %s", data + 1);
  int len = strlen(data + 1);
  scanf("%d", &k);
  if (len % k || k > len) {
    puts("NO");
    return 0;
  }
  int cut = len / k;
  bool flag = 1;
  for (int i = 1; i <= len; i += cut) {
    int l = i, r = i + cut - 1;
    while (l < r) {
      if (data[l] != data[r]) {
        flag = 0;
        break;
      }
      ++l, --r;
    }
    if (!flag) break;
  }
  if (!flag)
    puts("NO");
  else
    puts("YES");
  return 0;
}

```