#include <bits/stdc++.h>
#pragma comment(linker, ”/ STACK : 36777216“)
int ddx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
using namespace std;
int main() {
  int n;
  string s;
  cin >> s;
  int k;
  cin >> k;
  if (s.size() % k) {
    cout << "NO" << endl;
    return 0;
  }
  int len = s.size() / k;
  string first = "";
  for (int _n((s.size()) - 1), i(0); i <= _n; ++i) {
    first += s[i];
    if (first.size() == len) {
      for (int _n((len)-1), j(0); j <= _n; ++j)
        if (first[j] != first[len - j - 1]) {
          cout << "NO" << endl;
          return 0;
        }
      first = "";
    }
  }
  cout << "YES" << endl;
  cin >> n;
  return 0;
}

### Prompt

Develop a solution in cpp 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>
#pragma comment(linker, ”/ STACK : 36777216“)
int ddx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int ddy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
using namespace std;
int main() {
  int n;
  string s;
  cin >> s;
  int k;
  cin >> k;
  if (s.size() % k) {
    cout << "NO" << endl;
    return 0;
  }
  int len = s.size() / k;
  string first = "";
  for (int _n((s.size()) - 1), i(0); i <= _n; ++i) {
    first += s[i];
    if (first.size() == len) {
      for (int _n((len)-1), j(0); j <= _n; ++j)
        if (first[j] != first[len - j - 1]) {
          cout << "NO" << endl;
          return 0;
        }
      first = "";
    }
  }
  cout << "YES" << endl;
  cin >> n;
  return 0;
}

```