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.*;

public class Main {
    public static void main(String[] args) {
        Main main = new Main();
        main.run();
    }
    
    // 変数一覧
    String s;
    char[] ch;
    char[] ch1;
    char[] ch2;

    public void run() {
        // 入力情報取得
        Scanner sc = new Scanner(System.in);
        s = sc.next();
        ch = s.toCharArray();
        sc.close();

        // 解答処理
        solve();
    }
    
    private void solve() {
        int n = ch.length;
        boolean c1 = true;
        boolean c2 = true;

        ch1 = new char[(n-1)/2];
        ch2 = new char[(n-1)/2];
        for (int i = 0; i < (n - 1) / 2; i++) {
            ch1[i] = ch[i];
        }
        for (int i = 0; i <= n - (n + 3) / 2; i++) {
            ch2[i] = ch[i - 1 + ((n + 3) / 2)];
        }
        for (int i = 0; i < ch1.length / 2; i++) {
            if (ch1[i] != (ch1[ch1.length - i-1])) c1 = false;
        }
        for (int i = 0; i < ch2.length / 2; i++) {
            if (ch2[i] != (ch2[ch2.length - i-1])) c2 = false;
        }
        if (c1&&c2) {
            System.out.println("Yes");
        } else {
            System.out.println("No");
        }
    }
}