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# (.NET Core 3.1.201) (judge verdict: Wrong Answer)

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

namespace test
{
    class MainClass
    {
        public static void Main(string[] args)
        {

            //問題160

            string s = Console.ReadLine();

            char[] mae = new char[(s.Length-1)/2];
            char[] ato = new char[(s.Length - 1) / 2];

            for (int i = 0; i < (s.Length - 1) / 2; i++)
            {
                mae[i] = s[i];
            }

            int j = 0;
            for (int i = (s.Length + 1)/2; i < s.Length; i++)
            {
                //Console.WriteLine(i);
                ato[j] = s[i];
                j++;
            }

            //foreach(var a in mae)
            //{
            //    Console.WriteLine(a);
            //}

            //foreach (var a in ato)
            //{
            //    Console.WriteLine(a);
            //}

            for (int i = 0; i < mae.Length / 2; i++)
            {
                if (mae[i] != mae[mae.Length - 1 - i])
                {
                    Console.WriteLine("No");
                    return;
                }
            }


            for (int i = 0; i < ato.Length / 2; i++)
            {
                if (ato[i] != ato[ato.Length - 1 - i])
                {
                    Console.WriteLine("No");
                    return;
                }
            }

            Console.WriteLine("Yes");



        }

    }





    
}