A string S of an odd length is said to be a strong palindrome if and only if all of the following conditions are satisfied:

* S is a palindrome.
* Let N be the length of S. The string formed by the 1-st through ((N-1)/2)-th characters of S is a palindrome.
* The string consisting of the (N+3)/2-st through N-th characters of S is a palindrome.



Determine whether S is a strong palindrome.

Constraints

* S consists of lowercase English letters.
* The length of S is an odd number between 3 and 99 (inclusive).

Input

Input is given from Standard Input in the following format:


S


Output

If S is a strong palindrome, print `Yes`; otherwise, print `No`.

Examples

Input

akasaka


Output

Yes


Input

level


Output

No


Input

atcoder


Output

No
import java.util.*;

public class Main {
    static Scanner sc = new Scanner(System.in);

    public static void main(String[] args) {
        String s = sc.next();
        String s0 = s.substring(0, (s.length() - 1) / 2);
        String s1 = s.substring((s.length() + 1) / 2);
        System.out.println(isKaibun(s0) && s0.equals(s1) ? "Yes" : "No");
    }

    static boolean isKaibun(String str) {
        return str.length() <= 1 ? true :
                str.charAt(0) == str.charAt(str.length() -1) ? isKaibun(str.substring(1, str.length() - 1)) : false;
    }
}
