#include <bits/stdc++.h>
using namespace std;
string str;
bool check(int lo, int hi) {
  while (lo <= hi) {
    if (str[lo] != str[hi]) return false;
    lo++;
    hi--;
  }
  return true;
}
string ans[2] = {"NO", "YES"};
int main() {
  int k;
  cin >> str >> k;
  if (str.size() % k != 0) {
    puts("NO");
  } else {
    int len = (int)str.size() / k;
    int lo = 0, hi = -1;
    bool ok = true;
    for (int i = len - 1; i < str.size(); i += len) {
      lo = hi + 1;
      hi = i;
      if (!check(lo, hi)) {
        ok = false;
        break;
      }
    }
    cout << ans[ok] << endl;
  }
  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>
using namespace std;
string str;
bool check(int lo, int hi) {
  while (lo <= hi) {
    if (str[lo] != str[hi]) return false;
    lo++;
    hi--;
  }
  return true;
}
string ans[2] = {"NO", "YES"};
int main() {
  int k;
  cin >> str >> k;
  if (str.size() % k != 0) {
    puts("NO");
  } else {
    int len = (int)str.size() / k;
    int lo = 0, hi = -1;
    bool ok = true;
    for (int i = len - 1; i < str.size(); i += len) {
      lo = hi + 1;
      hi = i;
      if (!check(lo, hi)) {
        ok = false;
        break;
      }
    }
    cout << ans[ok] << endl;
  }
  return 0;
}

```