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 abc159b
{
    class Program
    {
        static void Main(string[] args)
        {
            //1) input
            string input1 = Console.ReadLine();
            //==========================================================================================================================




            //2) output
            string output = "No";

            int n = input1.Length;
            string a = input1.Substring(0, (n - 1) / 2);
            string b = input1.Substring((n - 1) / 2 + 1, (n - 1) / 2);
            bool isPalindrome = a == string.Join("", b.Reverse());


            int nA = a.Length;
            int sizeA = 0;
            int indexA = 0;
            int tmpA = 0;
            if (nA % 2 == 0)
            {
                tmpA = nA;
                sizeA = tmpA / 2;
                indexA = tmpA / 2;

            }
            else
            {
                tmpA = nA - 1;
                sizeA = tmpA / 2;
                indexA = tmpA / 2 + 1;

            }

            string aA = a.Substring(0, sizeA);           
            string bA = a.Substring(indexA, sizeA);
            bool isPalindromeA = aA == string.Join("", bA.Reverse());


            int nB = a.Length;
            int sizeB = 0;
            int indexB = 0;
            int tmpB = 0;
            if (nA % 2 == 0)
            {
                tmpB = nB;
                sizeB = tmpB / 2;
                indexB = tmpB / 2;

            }
            else
            {
                tmpB = nB - 1;
                sizeB = tmpB / 2;
                indexB = tmpB / 2 + 1;

            }

            string aB = b.Substring(0, sizeB);
            string bB = b.Substring(indexB, sizeB);
            bool isPalindromeB = aB == string.Join("", bB.Reverse());



            if (isPalindrome && isPalindromeA && isPalindromeB)
            {
                output = "Yes";
            }


            Console.WriteLine(output);
        }
    }
}