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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author silviase
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        BStringPalindrome solver = new BStringPalindrome();
        solver.solve(1, in, out);
        out.close();
    }

    static class BStringPalindrome {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            String s = in.next();
            String t = s.substring(0, s.length() / 2);
            String u = s.substring(s.length() / 2 + 1);

            if (isKaibun(s) && isKaibun(t) && isKaibun(u)) {
                out.println("Yes");
            } else {
                out.println("No");
            }
        }

        boolean isKaibun(String s) {
            for (int i = 0; i < s.length(); i++) {
                if (s.charAt(i) != s.charAt(s.length() - 1 - i)) {
                    return false;
                }
            }
            return true;
        }

    }
}