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.*;
    import java.io.*;

    public class Main {

    void solve(InputReader in, PrintWriter out) {
        char[] arr = in.next().toCharArray();
        int N = arr.length;
        for (int i = 0, j = (N-1)/2 - 1; i < j; ++i, --j) {
            if (arr[i] != arr[j]) {
                out.println("No");
                return;
            }
        }

        for (int i = N/2 + 1, j = N - 1; i < j; ++i, --j) {
            
            if (arr[i] != arr[j]) {
                out.println("No");
                return;
            }
        }

        out.println("Yes");
    }

    

    public static void main(String[] args) throws Exception {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);
        Main solver = new Main();
        solver.solve(in, out);
        out.close();
    }

    static class InputReader {
        BufferedReader in;
        StringTokenizer st;

        public InputReader(InputStream is) {
        in = new BufferedReader(new InputStreamReader(is));
        }

        public String next() {
        try {
            while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
            }
            return st.nextToken();
        } catch (Exception e) {
            throw new RuntimeException(e);
        }
        }

        public int nextInt() {
        return Integer.parseInt(next());
        }

        public long nextLong() {
        return Long.parseLong(next());
        }
    }
    }