#include <bits/stdc++.h>
using namespace std;
const int maxn = 1200 + 10;
char s[maxn];
int a[maxn];
int main() {
  while (~scanf("%s", s)) {
    memset(a, 0, sizeof(a));
    int len = strlen(s);
    int cnt = 0;
    int m;
    scanf("%d", &m);
    if (len % m == 0) {
      int cur = len / m;
      int sign = 1;
      for (int i = 0; i < len; i += cur) {
        for (int j = i, k = i + cur - 1; j < k; j++, k--) {
          if (s[j] != s[k]) {
            sign = 0;
            break;
          }
        }
        if (sign == 0) break;
      }
      if (sign) cnt = 1;
    }
    if (cnt)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
const int maxn = 1200 + 10;
char s[maxn];
int a[maxn];
int main() {
  while (~scanf("%s", s)) {
    memset(a, 0, sizeof(a));
    int len = strlen(s);
    int cnt = 0;
    int m;
    scanf("%d", &m);
    if (len % m == 0) {
      int cur = len / m;
      int sign = 1;
      for (int i = 0; i < len; i += cur) {
        for (int j = i, k = i + cur - 1; j < k; j++, k--) {
          if (s[j] != s[k]) {
            sign = 0;
            break;
          }
        }
        if (sign == 0) break;
      }
      if (sign) cnt = 1;
    }
    if (cnt)
      printf("YES\n");
    else
      printf("NO\n");
  }
  return 0;
}

```