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){
		try(Scanner sc = new Scanner(System.in)) {
			
			String s = sc.next();
			
			int width = (s.length() - 1)/2;
			
			boolean isStrong = checkCircle(s) && checkCircle(s.substring(0, width)) &&
					checkCircle(s.substring(width + 1, s.length()));
			
			System.out.println(isStrong ? "Yes" : "No");
		}
	}
	
	public static boolean checkCircle(String s) {
		
//		System.out.println(s);
		
		for(int i = 0  ; i < (s.length()-1) / 2 ; i++ ) {
//			System.out.println(s.charAt(i));
			if ( s.charAt(i) == s.charAt(s.length()-1-i)) {
				//ok
			}
			else {
				return false;
			}
		}
		return true;
	}

}