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: Accepted)

using System;

namespace abc159_b
{
    class Program
    {
        static void Main(string[] args)
        {
            string s = Console.ReadLine();
            int n = s.Length;

            Program p = new Program();
            if (!p.isKaibun(s))
            {
                Console.WriteLine("No");
                return;
            }

            int nFront = (n - 1) / 2;
            string sFront = s.Substring(0, nFront);
            if (!p.isKaibun(sFront))
            {
                Console.WriteLine("No");
                return;
            }

            int nBack = (n + 3) / 2;
            string sBack = s.Substring(nBack - 1);
            if (!p.isKaibun(sBack))
            {
                Console.WriteLine("No");
                return;
            }

            if (sFront == sBack)
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }
        }

        private bool isKaibun(string s)
        {
            bool kaibunHantei = true;
            int sKaibunHanteiLength = s.Length / 2;

            for (int i = 0;i < sKaibunHanteiLength; i++)
            {
                string sFront = s.Substring(i, 1);
                string sBack = s.Substring((s.Length-i-1),1);
                if(sFront == sBack)
                {
                    continue;
                }
                else
                {
                    kaibunHantei = false;
                }
            }

            return kaibunHantei;
        }
    }
}