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 Java (OpenJDK 11.0.6) (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		String string = sc.nextLine();
		Main main = new Main();

		if (main.chkPalindrome(string) && main.chkPalindrome(string.substring(0, (string.length() - 1)/2)) && main.chkPalindrome(string.substring((string.length() + 2)/2,string.length()))) {
			System.out.println("Yes");
		}else {
			System.out.println("No");
		}
	}

	public boolean chkPalindrome(String string) {
		int len = string.length();
		int right = len - 1;
		int left = 0;
		while (true) {
			if (len % 2 == 0 && right < left) {
				break;
			}else if(len % 2 == 1 && right == left){
				break;
			}
			if (string.charAt(left) != string.charAt(right)) {
				return false;
			}
			left++;
			right--;
		}
		return true;
	}
}