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 s = sc.next();
		String[] sArray = s.split("");
		for(int i =0;i < (sArray.length -1)/2;i++) {
			if(!sArray[i].equals(sArray[sArray.length-1-i])) {
				System.out.println("No");
				return;
			}
		}
		String left = s.substring(0, (s.length()-1)/2);
		String[] leftArray = left.split("");
		for(int i =0;i < leftArray.length/2;i++) {
			if(!leftArray[i].equals(leftArray[leftArray.length-1-i])) {
				System.out.println("No");
				return;
			}
		}

		String right = s.substring(((s.length()+3)/2) -1);
		String[] rightArray = right.split("");
		for(int i =0;i < rightArray.length/2;i++) {
			if(!rightArray[i].equals(rightArray[rightArray.length-1-i])) {
				System.out.println("No");
				return;
			}
		}
		System.out.println("Yes");

	}
}