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[]) {
		Scanner can = new Scanner(System.in);
		String kaibun = can.next();
		char bunkai[] = new char[kaibun.length()];
		
		String zenhan = kaibun.substring(0, (kaibun.length() - 1) / 2 );
		String kouhan = kaibun.substring((kaibun.length() - 1)/ 2 + 1);
		
		if(!zenhan.equals(kouhan)) {
			System.out.println("no");
			System.exit(0);
			
		}
		
		for(int i = 0; i < zenhan.length(); i++) {
			bunkai[i] = zenhan.charAt(i);
			
		}
		
		for(int i = 0; i < (zenhan.length() - 1) / 2; i++) {
			if(bunkai[i] != bunkai[zenhan.length() - 1 - i]) {
				System.out.println("no");
				System.exit(0);
				
			}
			
		}
		
		System.out.println("yes");
		
		
	}
	
}