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;

namespace AtCoder20200322
{
    class Program
    {
        static void Main(string[] args)
        {
            // TheNumberOfEvenPairs();
            StringPalindrome();
            // MaxVolume();
            // BannedK();
        }
        
        public static void TheNumberOfEvenPairs()
        {
            String[] inputs = Console.ReadLine().Split(' ');
            int n = int.Parse(inputs[0]);
            int m = int.Parse(inputs[1]);

            int sum = nCm(n, 2) + nCm(m, 2);
            Console.WriteLine(sum);
        }

        public static int nCm(int n, int m)
        {
            if (m == 0) return 1;
            if (n == 0) return 0;
            return n * nCm(n - 1, m - 1) / m;
        }

        public static void MaxVolume()
        {
            double l = double.Parse(Console.ReadLine());
            double vol = Math.Pow(l, 3) / (double)27;
            Console.WriteLine(vol);
        }

        public static void BannedK()
        {
            int n = int.Parse(Console.ReadLine());
            String[] input_a = Console.ReadLine().Split(' ');
            int[] a = new int[n];
            Dictionary<int, int> dic = new Dictionary<int, int>();
            for (int i=0; i<n; i++)
            {
                a[i] = int.Parse(input_a[i]);
                if (dic.ContainsKey(a[i]))
                {
                    dic[a[i]] += 1;
                }
                else
                {
                    dic.Add(a[i], 1);
                }
            }
            Dictionary<int, int> sum_dic = new Dictionary<int, int>();
            Dictionary<int, int> reduced_sum_dic = new Dictionary<int, int>();
            int sum = 0;
            
            foreach(int key in dic.Keys)
            {
                int sum_key = nCm(dic[key], 2);
                sum_dic.Add(key,sum_key);
                reduced_sum_dic.Add(key, nCm(dic[key] - 1, 2));
                sum += sum_key;
            }

            // Result
            foreach(int i in a)
            {
                int res = sum - sum_dic[i] + reduced_sum_dic[i];
                Console.WriteLine(res);
            }
        }

        public static void StringPalindrome()
        {
            String s = Console.ReadLine();
            int n = s.Length;
            char[] cs = new char[n];
            char[] c_former = new char[(n - 1) / 2];
            char[] c_latter = new char[(n - 1) / 2];
            int i = 0;
            foreach(char c in s)
            {
                cs[i] = c;
                i++;
            }
            bool isKaibun = true;
            for(int j=0; j<(n-1)/2; j++)
            {
                isKaibun = isKaibun && (cs[j] == cs[n - 1 - j]);
                c_former[j] = cs[j];
                c_latter[(n - 1) / 2 - 1- j] = cs[n - 1 - j];
            }
            if(!isKaibun)
            {
                Console.WriteLine("No");
                return;
            }
            for(int j=0; j<c_former.Length; j++)
            {
                isKaibun = isKaibun && (c_former[j] == c_former[c_former.Length - 1 - j]) && (c_latter[j] == c_latter[c_latter.Length - 1 - j]);   
            }
            if(isKaibun)
            {
                Console.WriteLine("Yes");
            }
            else
            {
                Console.WriteLine("No");
            }


        }
    }
}