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.*;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        String s = sc.next();
        boolean ans = check(s);
        System.out.println(ans ? "Yes" : "No");
        sc.close();

    }

    private static boolean check(String s) {
        if(s.length() == 1) return true;
        boolean ret = true;
        ret = isKaibun(s);
        if(ret){
            ret = check(s.substring(0, s.length()/2));
        }
        if(ret){
            ret = check(s.substring(s.length()/2 + s.length()%2, s.length()));
        }
        return ret;
    }

    private static boolean isKaibun(String s) {
        for (int i = 0; i < s.length()/2; i++) {
            if(s.charAt(i) != s.charAt(s.length()-1-i)){
                return false;
            }
        }
        return true;
    }

}