#include <bits/stdc++.h>
char lin[1024];
int N, K, lenSection;
bool isPalin(int st, int fn) {
  for (int i = 0; i < lenSection / 2; i++) {
    if (lin[st + i] != lin[fn - i]) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%s\n%d", lin, &K);
  N = strlen(lin);
  if (N % K != 0) {
    printf("NO\n");
    return 0;
  }
  lenSection = N / K;
  for (int i = 0; i < K; i++) {
    if (!isPalin(i * lenSection, (i + 1) * lenSection - 1)) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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>
char lin[1024];
int N, K, lenSection;
bool isPalin(int st, int fn) {
  for (int i = 0; i < lenSection / 2; i++) {
    if (lin[st + i] != lin[fn - i]) {
      return false;
    }
  }
  return true;
}
int main() {
  scanf("%s\n%d", lin, &K);
  N = strlen(lin);
  if (N % K != 0) {
    printf("NO\n");
    return 0;
  }
  lenSection = N / K;
  for (int i = 0; i < K; i++) {
    if (!isPalin(i * lenSection, (i + 1) * lenSection - 1)) {
      printf("NO\n");
      return 0;
    }
  }
  printf("YES\n");
  return 0;
}

```