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;

class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		String s = sc.next();
		int n = s.length();
		
		// 条件１
		if (!isKaibun(s)) {
			System.out.println("No");
			sc.close();
			return;
		}
		
		// 条件２
		String sub1 = s.substring(0, (n - 1) / 2);
		if (!isKaibun(sub1)) {
			System.out.println("No");
			sc.close();
			return;
		}
		
		// 条件３
		String sub2 = s.substring((n + 3) / 2 - 1, n);
		if (!isKaibun(sub2)) {
			System.out.println("No");
			sc.close();
			return;
		}
		
		System.out.println("Yes");
		sc.close();
	}
	
	static boolean isKaibun(String s) {
		char[] c = s.toCharArray();
		boolean isKaibun = true;
		int len = c.length;
		for (int i = 0; i < len / 2; i++) {
			if (c[i] != c[len - 1 - i]) {
				isKaibun = false;
				break;
			}
		}
		return isKaibun;
	}
}