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 sc = new Scanner(System.in);
    String A = sc.next();
    int b = A.length();
    if(b >= 3 && b <= 99 && b % 2 != 0){
    int c = (A.length()-1)/2;
    String B = A.substring(0,c);
    int d = (A.length() + 3)/2;
    String C = A.substring(d-1,A.length());
    //System.out.println(B+" "+C);
    int e = 0;
    for(int a = 0; a < B.length();a++){
      char D = B.charAt(a);
      char E = B.charAt(B.length()-1-a);
      if(D != E){
        e = 1;
      }
    }
    for(int a = 0; a < C.length();a++){
      char D = C.charAt(a);
      char E = C.charAt(C.length()-1-a);
      if(D != E){
        e = 1;
      }
    }
    if(e == 0){
      System.out.println("Yes");
    }
    else{
      System.out.println("No");
    }
  }
  }}