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: Wrong Answer)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		// 読み込み
		Boolean flag1 = false;
		Boolean flag2 = false;
		Boolean flag3 = false;
		
		
		Scanner sc = new Scanner(System.in);
		String S = sc.next();
		int N = S.length();
		
		StringBuffer sb = new StringBuffer(S);
		StringBuffer sbr = sb.reverse();
		if(sb.toString().equals(sbr.toString())) {
			flag1 = true;
		}
		
		String parse = S.substring(0, (N -1)/2);
		sb = new StringBuffer(parse);
		sbr = sb.reverse();
		if(sb.toString().equals(sbr.toString())) {
			flag2 = true;
		}
		//System.out.println(sb + " " + sbr);
	
		parse = S.substring((N+3)/2-1, N);
		sb = new StringBuffer(parse);
		sbr = sb.reverse();
		if(sb.toString().equals(sbr.toString())) {
			flag3 = true;
		}
		//System.out.println(sb + " " + sbr);
		
		if(flag1 == true && flag2 == true && flag3 == true) {
			System.out.println("Yes");
		}
		else {
			System.out.println("No");
		}
	}
}