#include <bits/stdc++.h>
using namespace std;
const int MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
const int MXN = 1000010;
const int C = 262144;
const int INF = 1000000001;
char str[MXN];
int n, k;
bool solve() {
  if (n % k) return 0;
  int p = n / k;
  for (int i = (1); i <= (k); i++)
    for (int j = (1); j <= (p); j++)
      if (str[(i - 1) * p + j] != str[i * p - j + 1]) return 0;
  return 1;
}
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  scanf("%d", &k);
  printf(solve() ? "YES\n" : "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 MAX_BUF_SIZE = 16384;
char BUFOR[MAX_BUF_SIZE];
int BUF_SIZE, BUF_POS;
char ZZZ;
const int MXN = 1000010;
const int C = 262144;
const int INF = 1000000001;
char str[MXN];
int n, k;
bool solve() {
  if (n % k) return 0;
  int p = n / k;
  for (int i = (1); i <= (k); i++)
    for (int j = (1); j <= (p); j++)
      if (str[(i - 1) * p + j] != str[i * p - j + 1]) return 0;
  return 1;
}
int main() {
  scanf("%s", str + 1);
  n = strlen(str + 1);
  scanf("%d", &k);
  printf(solve() ? "YES\n" : "NO\n");
  return 0;
}

```