#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, a;
  int k, x = 0;
  cin >> s >> k;
  int len = s.length();
  if (len % k != 0) {
    cout << "NO" << '\n';
    return 0;
  }
  vector<string> word;
  k = len / k;
  for (int i = 0; i < len; i++) {
    a += s[i];
    x++;
    if (x == k) {
      word.push_back(a);
      a = "";
      x = 0;
    }
  }
  bool flag = true;
  for (int i = 0; i < word.size(); i++) {
    len = word[i].length();
    for (int j = 0, k = len - 1; j < k; j++, k--) {
      if (word[i][j] != word[i][k]) {
        flag = false;
        break;
      }
    }
    if (!flag) {
      break;
    }
  }
  if (flag) {
    cout << "YES" << '\n';
  } else {
    cout << "NO" << '\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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  string s, a;
  int k, x = 0;
  cin >> s >> k;
  int len = s.length();
  if (len % k != 0) {
    cout << "NO" << '\n';
    return 0;
  }
  vector<string> word;
  k = len / k;
  for (int i = 0; i < len; i++) {
    a += s[i];
    x++;
    if (x == k) {
      word.push_back(a);
      a = "";
      x = 0;
    }
  }
  bool flag = true;
  for (int i = 0; i < word.size(); i++) {
    len = word[i].length();
    for (int j = 0, k = len - 1; j < k; j++, k--) {
      if (word[i][j] != word[i][k]) {
        flag = false;
        break;
      }
    }
    if (!flag) {
      break;
    }
  }
  if (flag) {
    cout << "YES" << '\n';
  } else {
    cout << "NO" << '\n';
  }
  return 0;
}

```