Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

akasaka

Sample Output 1

Yes

S is akasaka.

The string formed by the 1-st through the 3-rd characters is aka.

The string formed by the 5-th through the 7-th characters is aka.
All of these are palindromes, so S is a strong palindrome.

Sample Input 2

level

Sample Output 2

No

Sample Input 3

atcoder

Sample Output 3

No

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

public class Main
{
	public static void main(String[] args)
	{
		Scanner sc = new Scanner(System.in);
		String str = sc.nextLine();
		sc.close();
		int n = str.length();

        StringBuffer sb2 = new StringBuffer(str);
		String dst2 = sb2.reverse().toString();
		if (str.compareTo(dst2) != 0) {
			System.out.println("No");
			return;
		}

		String sub1 = str.substring(0,(n-1)/2);
		StringBuffer sb = new StringBuffer(sub1);
		String dst = sb.reverse().toString();
		if (sub1.compareTo(dst) != 0) {
			System.out.println("No");
			return;
		}
		sub1 = str.substring((n+3)/2-1, n);
        sb = new StringBuffer(sub1);
		dst = sb.reverse().toString();
		if (sub1.compareTo(dst) != 0) {
			System.out.println("No");
			return;
		}



		System.out.println("Yes");
	}
}