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.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCoder
{
	class Program
	{
		static void Main(string[] args)
		{
			var s = Console.ReadLine();
			if(!IsPalindrome(s))
			{
				Console.WriteLine("No");
				return;
			}
			if(!IsPalindrome(s.Substring(0, (s.Length - 1) / 2)))
			{
				Console.WriteLine("No");
				return;
			}
			if (!IsPalindrome(s.Substring((s.Length + 3) / 2 - 1, (s.Length - 1) / 2)))
			{
				Console.WriteLine("No");
				return;
			}
			Console.WriteLine("Yes");
		}

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