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.Scanner;

public class Main{
  public static void main(String[] args){
    Scanner scanner = new Scanner(System.in);
    String S = scanner.next();
    
    int N = S.length();
    String S1 = S.substring(0,(N - 1) / 2);
    String S2 = S.substring((N + 3) / 2 - 1,N);
    
    String answer = "No";
    if(k(S,N) && k(S1,S1.length()) && k(S2,S2.length())){
      answer = "Yes";
    }
    
    System.out.println(answer);
  }
  
  public static boolean k(String str,int length){
    for(int i = 0;i < (length + 1) / 2;i++){
         /*System.out.println(
          i + "," + 
          str.charAt(i) + "," + str.charAt(length - i - 1));
      */
      if(str.charAt(i) != str.charAt(length - i - 1)){
        return false;
      }
    }
    return true;
  }
}