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: Accepted)

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

namespace AtCoder {
    class Program {
        static void Main(string[] args) {
            string buf = Console.ReadLine();
            char[] sentence = buf.ToCharArray();
            int n = sentence.Length;

            if (n % 2 == 0) {
                Console.WriteLine("No");
                return;
            }

/*            if ((n - 1) / 2 % 2 == 0) {
                Console.WriteLine("No");
                return;
            }*/

            char[] test1 = new char[n / 2];
            char[] test2 = new char[n / 2];

            Array.Copy(sentence, 0, test1, 0, n / 2);
            Array.Copy(sentence, n / 2 + 1, test2, 0, n / 2);

            Array.Reverse(test2);

            //Console.WriteLine(new string(test1));
            //Console.WriteLine(new string(test2));


            string test1Str = new string(test1);
            string test2Str = new string(test2);

            //1. Sは回文かどうか
            if (test1Str != test2Str) {
                Console.WriteLine("No");
                return;
            }

            char[] test3 = test1;
            Array.Reverse(test3);
            string test3Str = new string(test3);

            //2. 前半分が回分かどうか
            if (test1Str != test3Str) {
                Console.WriteLine("No");
                return;
            }

            char[] test4 = test2;
            char[] test5 = test4;
            Array.Reverse(test5);

            string test4Str = new string(test4);
            string test5Str = new string(test5);

            //Console.WriteLine(new string(test4));

            if (test4Str != test5Str) {
                Console.WriteLine("No");
                return;
            }

            Console.WriteLine("Yes");
        }

    }

}