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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>

int main()
{
	char inputS[99];
	long inputLength;
	int isPalindrome = 1;

	scanf("%s", inputS);
	inputLength = strlen(inputS);

	// Check for words with odd number of letters and 
	if (inputLength % 2 > 0)
	{
		for (int i = 0; i < inputLength / 2; i++)
		{
			// If there is difference between right and left side. If there is difference then prints 'No' and terminates function
			if (inputS[i] != inputS[inputLength / 2 + i + 1])
			{
				isPalindrome = 0;
				break;
			}
		}
	}
	else if (inputLength % 2 == 0)
	{
		// Code for words with even number of letters
		for (int i = 0; i < inputLength / 2; i++)
		{
			if (inputS[i] != inputS[inputLength / 2 + i])
			{
				isPalindrome = 0;
				break;
			}
		}
	}

	if (isPalindrome)
	{
		printf("Yes \n");
	}
	else
	{
		printf("No \n");

	}

	return 0;
}