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: Wrong Answer)

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

int Is_Palindrome(char *str, int length)
{
	int count = 0;
	int i;

	count = length % 2;

	for(i = 0; i < count; i++)
	{
		if( *(str + i) != *(str + length - 1 - i) )
		{
			return 0;
		}
	}

	count = ((length - 1) / 2) % 2;

	for(i = 0; i < count; i++)
	{
		if( *(str + i) != *(str + (length - 1)/2 - 1 - i) )
		{
			return 0;
		}
	}

	for(i = 0; i < count; i++)
	{
		if( *(str + (length + 3)/2 - 1 + i) != *(str + length - 1 - i) )
		{
			return 0;
		}
	}

	return 1;
}

int main()
{
	char S[128];
	int flag = 0;

	scanf("%s", S);

	if(strlen(S) % 4 == 3)
	{
		if(strlen(S) == 3 && (*(S) == *(S + strlen(S) - 1) ))
	        {
        	        flag = 1;
        	}
		else if(Is_Palindrome(S, strlen(S)) == 1)
		{
			flag = 1;
		}
		else
		{
			flag = 0;
		}
	}
	else
	{
		printf("No");
		return 0;
	}

	if(flag == 1)
	{
		printf("Yes");
	}
	else
	{
		printf("No");
	}

	return 0;
}