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.IO.Pipes;
using System.Linq;
using System.Text;

namespace cswork
{
    class Program
    {
        class Scanner
        {
            string[] s;
            int i;

            char[] cs = new char[] { ' ' };

            public Scanner()
            {
                s = new string[0];
                i = 0;
            }

            public string next()
            {
                if (i < s.Length) return s[i++];
                string st = Console.ReadLine();
                while (st == "") st = Console.ReadLine();

                s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
                i = 0;
                return next();
            }

            public int nextInt()
            {
                return int.Parse(next());
            }

            public long nextLong()
            {
                return long.Parse(next());
            }

            public double nextDouble()
            {
                return double.Parse(next());
            }
        }

        public Program() { }

        static void Main(string[] args)
        {
            new Program().Omame();
        }

        Scanner cin;
        const int MAXIN = 1024;
        bool[] used = new bool[MAXIN];
        int[] perm = new int[MAXIN];

        List<int[]> permList = new List<int[]>();

        public void Permutation(int pos, int n, int length)
        {
            if (pos == n)
            {
                permList.Add((int[])perm.Clone());
            }

            for (int i = 0; i < n; i++)
            {
                if (!used[i])
                {
                    perm[pos] = i;
                    used[i] = true;
                    Permutation(pos + 1, n, length);
                    used[i] = false;
                }
            }
        }

        void Omame()
        {
            cin = new Scanner();
            string S = cin.next();

            if (!isKaibun(S))
            {
                Console.WriteLine("No");
                return;
            }

            if (!isKaibun(S.Substring(0, (S.Length - 1) / 2)))
            {
                Console.WriteLine("No");
                return;
            }

            if (!isKaibun(S.Substring((S.Length + 3) / 2 - 1)))
            {
                Console.WriteLine("No");
                return;
            }

                Console.WriteLine("Yes");
        }

        bool isKaibun(string s)
        {
            for (int i = 0; i < s.Length / 2; i++)
            {
                if (s[i] != s[s.Length - 1 - i])
                {
                    return false;
                }
            }
            return true;
        }
    }
}