#include <bits/stdc++.h>
using namespace std;
int main() {
  string bag;
  int k;
  cin >> bag >> k;
  int wl;
  if (bag.size() % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int flag = 0;
  string pal1, pal2;
  wl = bag.size() / k;
  for (int i = 0; i < bag.size(); i++) {
    if (i % wl == 0) {
      pal2 = pal1;
      reverse(pal2.begin(), pal2.end());
      if (pal1 != pal2) {
        flag = 1;
      }
      pal1.clear();
    }
    pal1.push_back(bag[i]);
  }
  pal2 = pal1;
  reverse(pal2.begin(), pal2.end());
  if (pal1 != pal2) {
    flag = 1;
  }
  if (flag) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int main() {
  string bag;
  int k;
  cin >> bag >> k;
  int wl;
  if (bag.size() % k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int flag = 0;
  string pal1, pal2;
  wl = bag.size() / k;
  for (int i = 0; i < bag.size(); i++) {
    if (i % wl == 0) {
      pal2 = pal1;
      reverse(pal2.begin(), pal2.end());
      if (pal1 != pal2) {
        flag = 1;
      }
      pal1.clear();
    }
    pal1.push_back(bag[i]);
  }
  pal2 = pal1;
  reverse(pal2.begin(), pal2.end());
  if (pal1 != pal2) {
    flag = 1;
  }
  if (flag) {
    cout << "NO" << endl;
  } else {
    cout << "YES" << endl;
  }
  return 0;
}

```