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 {
public static void main(String[] args){
	Scanner sc = new Scanner(System.in);
	String s = sc.next();
  	int l = s.length();
  	if(l/2==0) System.exit(0);
  	if((l<3)||(99<l)) System.exit(0);
  
  	if (kaibunCheck(s)) {
      if (kaibunCheck(s.substring(0,(l-1)/2))) {
        if (kaibunCheck(s.substring((l+3)/2-1,l))) {
        	System.out.println("Yes");
          	System.exit(0);
      	}
      }
    }
      System.out.println("No");
	}
  
public static boolean kaibunCheck(String str){
  	int len = str.length();
  	char[] c = new char[len];
  	for(int i=0;i<len;i++){
      c[i] = str.charAt(i);
    }
  	for(int j=0;j<len;j++){
      if(c[j]!=c[len-j-1]) return false;
    }
  	return true;
}
}