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 C# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;

namespace AtCoder
{
    class Program
    {
        static void Main(string[] args)
        {
            //[summary]C - Maximum Volume
            var S = Console.ReadLine();

            if (IsKaibun(S))
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }

        static bool IsKaibun(string S)
        {
            if (IsMatch(S))
            {
                int n = S.Length / 2;

                if (IsMatch(S.Substring(0, n)))
                {
                    if (IsMatch(S.Substring(n + 1, n)))
                    {
                        return true;
                    }
                }
            }

            return false;
        }

        static bool IsMatch(string s)
        {
            if (s.Length % 2 == 1)
            {
                var n = s.Length / 2;
                var left = s.Substring(0, n);
                var right = s.Substring(n + 1, n);

                if (left == right)
                {
                    return true;
                }
            }

            return false;
        }
    }
}