#include <bits/stdc++.h>
using namespace std;
bool is(string s) {
  int len = s.size();
  int lp = 0, rp = len - 1;
  while (lp <= rp) {
    if (s[lp] == s[rp]) {
      lp++;
      rp--;
    } else {
      return false;
    }
  }
  return true;
}
int main() {
  string s = "";
  cin >> s;
  int k, count = 0;
  scanf("%d", &k);
  int n = s.size();
  if (n % k != 0) {
    printf("NO\n");
    return 0;
  }
  int len = n / k;
  for (int i = 0; i < n; i += len) {
    string ans = "";
    for (int j = 0; j < len; j++) {
      ans += s[i + j];
    }
    if (is(ans)) {
      count++;
    }
  }
  if (count == k) {
    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;
bool is(string s) {
  int len = s.size();
  int lp = 0, rp = len - 1;
  while (lp <= rp) {
    if (s[lp] == s[rp]) {
      lp++;
      rp--;
    } else {
      return false;
    }
  }
  return true;
}
int main() {
  string s = "";
  cin >> s;
  int k, count = 0;
  scanf("%d", &k);
  int n = s.size();
  if (n % k != 0) {
    printf("NO\n");
    return 0;
  }
  int len = n / k;
  for (int i = 0; i < n; i += len) {
    string ans = "";
    for (int j = 0; j < len; j++) {
      ans += s[i + j];
    }
    if (is(ans)) {
      count++;
    }
  }
  if (count == k) {
    printf("YES\n");
  } else {
    printf("NO\n");
  }
  return 0;
}

```