#include <bits/stdc++.h>
using namespace std;
bool pali(string s) {
  int n = s.length();
  for (int i = 0; i <= n / 2; i++) {
    if (s[i] != s[n - 1 - i]) return false;
  }
  return true;
}
bool fun(vector<string> v) {
  for (auto x : v) {
    if (!pali(x)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  int q;
  cin >> q;
  vector<string> v;
  int len = (str.length()) / q;
  if (str.length() < q || ((str.length()) % q != 0)) {
    cout << "NO";
    return 0;
  }
  for (int i = 0; i < str.length(); i += len) {
    string s;
    for (int j = i; j < i + len; j++) s += str[j];
    v.push_back(s);
  }
  if (fun(v))
    cout << "YES";
  else
    cout << "NO";
}

### 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;
bool pali(string s) {
  int n = s.length();
  for (int i = 0; i <= n / 2; i++) {
    if (s[i] != s[n - 1 - i]) return false;
  }
  return true;
}
bool fun(vector<string> v) {
  for (auto x : v) {
    if (!pali(x)) return false;
  }
  return true;
}
int main() {
  string str;
  cin >> str;
  int q;
  cin >> q;
  vector<string> v;
  int len = (str.length()) / q;
  if (str.length() < q || ((str.length()) % q != 0)) {
    cout << "NO";
    return 0;
  }
  for (int i = 0; i < str.length(); i += len) {
    string s;
    for (int j = i; j < i + len; j++) s += str[j];
    v.push_back(s);
  }
  if (fun(v))
    cout << "YES";
  else
    cout << "NO";
}

```