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.*;


class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		String S = sc.next();
		int N = S.length();
		if(N <3){
			System.out.println("No");
			return;
		}
		
		List<String> s1 = new ArrayList();
		List<String> aa = new ArrayList();
		List<String> bb = new ArrayList();
		for(int i = 0;i < N;i++){
			s1.add(S.substring(i,i+1));
			if(i < (N-1)/2){
				aa.add(S.substring(i,i+1));
			}
			if(i > (N+3)/2){
				bb.add(S.substring(i,i+1));
			}
		}
		Collections.reverse(aa);
		Collections.reverse(bb);
		
		for(int i = 0;i < aa.size();i++){
			if(!aa.get(i).equals(s1.get(i))){
				System.out.println("No");
				return;
			}
		}
		for(int i = 0;i < bb.size();i++){
			if(!bb.get(i).equals(s1.get((i+3)/2+1))){
				System.out.println("No");
				return;
			}
		}
		
		
		System.out.println("Yes");
	}

}