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.Diagnostics;
using System.Linq;

namespace AtCoder.B
{
    internal static class CR
    {
        public static string GetText() { return Console.ReadLine(); }
        public static List<string> GetTexts() { return Console.ReadLine().Split(' ').ToList(); }
        public static long GetLong() { return long.Parse(Console.ReadLine()); }
        public static List<long> GetLongs() { return Console.ReadLine().Split(' ').Select(x => long.Parse(x)).ToList(); }
    }

    public class Program
    {
        public static string S { get; private set; }
        public static int N { get; private set; }

        public static void Main() { var r = GetResult(); Debug.WriteLine(r); Console.Write(r); }

        private static object GetResult()
        {
            S = CR.GetText();
            N = S.Length;

            if (!IsKaibun(S))
            {
                return "No";
            }

            var s1 = S.Substring(0, (N - 1) / 2);
            if (!IsKaibun(s1))
            {
                return "No";
            }
            var s2 = S.Substring((N + 3) / 2 - 1);
            if (!IsKaibun(s2))
            {
                return "No";
            }

            return "Yes";
        }

        private static bool IsKaibun(string s)
        {
            var harfIndex = s.Length / 2;
            for (int i = 0; i < harfIndex; i++)
            {
                if (s[i] != s[s.Length - 1 - i])
                {
                    return false;
                }
            }

            return true;
        }
    }
}