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 sc = new Scanner(System.in);
		char[] S = sc.next().toCharArray();
		sc.close();
		boolean ok = true;
		int N = S.length;
		// Sは回文であるか
		for(int i = 0; i < N; i++) {
			ok &= S[i] == S[N - i - 1];
		}
		// Sの1文字目から(N - 1)/2文字目まで(両端含む)からなる文字は回文であるか
		for(int i = 0; i < (N - 1) / 2; i++) {
			ok &= S[i] == S[(N - 1) / 2 - i - 1];
		}
		// Sの(N + 3)/2文字目からN文字目まで(両端含む)からなる文字は回文であるか
		// ※Sが回分であることが条件のため、Sの(N + 3)/2文字目移行の文字と1文字目～からが回分であればOK
		for(int i = (N + 3) / 2 - 1; i < N; i++) {
			ok &= S[i] == S[N - i - 1];
		}
		System.out.println(ok ? "Yes" : "No");
	}
}