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.*;

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        sc.close();

        if (StringPalindrome(S)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
    public static boolean StringPalindrome(String S) {
        // 1.Sは回文
        // 2.Sの1～(S.length()-1)/2文字目までが回文
        // 3.Sの(S.length()+3)/2～S.length()文字目までが回文
        StringBuffer sb = new StringBuffer(S);
        StringBuffer sb1 = new StringBuffer(S.substring(0, (S.length()-1)/2));
        if (sb.toString().compareTo(sb.reverse().toString()) == 0) {
            if (sb1.toString().compareTo(sb1.reverse().toString()) == 0) {
                return true;
            }
        }
        return false;
    }
}