#include <bits/stdc++.h>
using namespace std;
bool check(vector<char> v) {
  int i, j = 0;
  for (i = 0; i < v.size() / 2; i++) {
    if (v[i] != v[v.size() - 1 - i]) {
      j++;
      break;
    }
  }
  if (j == 0)
    return true;
  else
    return false;
}
int main() {
  int n, i, j = 0, k;
  string s, s1;
  vector<char> v;
  cin >> s;
  cin >> n;
  if (s.length() % n == 0) {
    k = 0;
    while (k != s.length()) {
      for (i = k; i < k + s.length() / n; i++) {
        v.push_back(s[i]);
      }
      if (!check(v)) {
        j++;
        break;
      }
      k += s.length() / n;
      v.clear();
    }
    if (j == 0)
      cout << "YES";
    else
      cout << "NO";
  } else
    cout << "NO";
  return 0;
}

### Prompt

Generate 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;
bool check(vector<char> v) {
  int i, j = 0;
  for (i = 0; i < v.size() / 2; i++) {
    if (v[i] != v[v.size() - 1 - i]) {
      j++;
      break;
    }
  }
  if (j == 0)
    return true;
  else
    return false;
}
int main() {
  int n, i, j = 0, k;
  string s, s1;
  vector<char> v;
  cin >> s;
  cin >> n;
  if (s.length() % n == 0) {
    k = 0;
    while (k != s.length()) {
      for (i = k; i < k + s.length() / n; i++) {
        v.push_back(s[i]);
      }
      if (!check(v)) {
        j++;
        break;
      }
      k += s.length() / n;
      v.clear();
    }
    if (j == 0)
      cout << "YES";
    else
      cout << "NO";
  } else
    cout << "NO";
  return 0;
}

```