#include <bits/stdc++.h>
char S[1010] = {0};
int main(void) {
  int k, Slen, ItemLen, flg = 1;
  scanf("%s", S);
  scanf("%d", &k);
  Slen = strlen(S);
  if (Slen % k) {
    flg = 0;
    goto PrintResult;
  }
  ItemLen = Slen / k;
  for (int i = 0; i < k; i++) {
    int a = i * ItemLen;
    int b = (i + 1) * ItemLen - 1;
    while (a < b) {
      if (S[a++] != S[b--]) {
        flg = 0;
        goto PrintResult;
      }
    }
  }
PrintResult:
  if (flg)
    printf("Yes");
  else
    printf("No");
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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>
char S[1010] = {0};
int main(void) {
  int k, Slen, ItemLen, flg = 1;
  scanf("%s", S);
  scanf("%d", &k);
  Slen = strlen(S);
  if (Slen % k) {
    flg = 0;
    goto PrintResult;
  }
  ItemLen = Slen / k;
  for (int i = 0; i < k; i++) {
    int a = i * ItemLen;
    int b = (i + 1) * ItemLen - 1;
    while (a < b) {
      if (S[a++] != S[b--]) {
        flg = 0;
        goto PrintResult;
      }
    }
  }
PrintResult:
  if (flg)
    printf("Yes");
  else
    printf("No");
  return 0;
}

```