#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k = 0;
  string s;
  cin >> s >> n;
  vector<char> v, vv;
  int len = s.size();
  if (len % n != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int cnt = 0, c = 0, j;
  int m = len / n;
  for (int i = 0; i < n; i++) {
    v.clear();
    vv.clear();
    for (j = c; j < len; j++) {
      v.push_back(s[j]);
      vv.push_back(s[j]);
      cnt++;
      c++;
      if (cnt == m) break;
    }
    cnt = 0;
    int pnt = 0;
    reverse(vv.begin(), vv.end());
    int ln = v.size();
    for (int i = 0; i < ln; i++) {
      if (v[i] == vv[i]) pnt++;
      if (pnt == ln) {
        k++;
      }
    }
    v.clear();
    vv.clear();
  }
  if (k == n) {
    k = 0;
    cout << "YES" << endl;
  } else {
    k = 0;
    cout << "NO" << 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;
int main() {
  int n, k = 0;
  string s;
  cin >> s >> n;
  vector<char> v, vv;
  int len = s.size();
  if (len % n != 0) {
    cout << "NO" << endl;
    return 0;
  }
  int cnt = 0, c = 0, j;
  int m = len / n;
  for (int i = 0; i < n; i++) {
    v.clear();
    vv.clear();
    for (j = c; j < len; j++) {
      v.push_back(s[j]);
      vv.push_back(s[j]);
      cnt++;
      c++;
      if (cnt == m) break;
    }
    cnt = 0;
    int pnt = 0;
    reverse(vv.begin(), vv.end());
    int ln = v.size();
    for (int i = 0; i < ln; i++) {
      if (v[i] == vv[i]) pnt++;
      if (pnt == ln) {
        k++;
      }
    }
    v.clear();
    vv.clear();
  }
  if (k == n) {
    k = 0;
    cout << "YES" << endl;
  } else {
    k = 0;
    cout << "NO" << endl;
  }
  return 0;
}

```