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 scn = new Scanner(System.in);

		String str = scn.next();

		int n = str.length();

		if (n % 2 == 0) {
			System.out.println("No");
		} else {

			int i = 0;
			int f = ((n - 1) / 2) - 1;

			boolean ini = ispalidrome(i, f, str);

			i = ((n + 3) / 2) - 1;
			f = n - 1;

			boolean fin = ispalidrome(i, f, str);

			if (ini && fin) {
				System.out.println("Yes");
			} else {
				System.out.println("No");
			}
		}

	}

	private static boolean ispalidrome(int i, int f, String str) {

		while (i <= f) {
			if (str.charAt(i) != str.charAt(f)) {
				return false;
			}
			i++;
			f--;
		}
		return true;
	}

}