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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		int n = s.length();
		if(kaibun(s, n)) {
			if(kaibun(s.substring(0, n / 2 + 1), n / 2)) {
				String s2 = s.substring((n + 3) / 2 - 1);
				if(kaibun(s2, s2.length())){
					System.out.println("Yes");
				}
			}
		}
		sc.close();
	}

	public static boolean kaibun(String s, int n) {
		for(int i = 0; i < n / 2; i++) {
			if(s.charAt(i) != s.charAt(n - i - 1)){
				System.out.println("No");
				return false;
			}
		}
		return true;
	}
}
