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 {
    private static final long MODVAL = 1000000007L;
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String S = sc.next();
        if(!isKaibun(S)) {
            System.out.println("No");
            return;
        }
        if(!isKaibun(S.substring(0, (S.length()-1)/2))) {
            System.out.println("No");
            return;
        }
        if(!isKaibun(S.substring((S.length()+3)/2 - 1, S.length()))) {
            System.out.println("No");
            return;
        }
        System.out.println("Yes");
     }

     static boolean isKaibun(String S) {
//        System.out.println(S);
        for(int i=0; i<S.length(); i++) {
            if(S.charAt(i) != S.charAt(S.length()-1-i)) return false;
         }
        return true;
     }
}