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>
#include <stdlib.h>

int main() {


	char s[100];
	scanf("%s", s);

	int length = strlen(s);
	int former = (length - 1) / 2;
	int latter = (length + 3) / 2;


	int f_judge = 0;
	int l_judge = 0;
	int i;

	for(i = 0; i < former / 2; i++) {
		if(s[i] == s[former - 1 - i]) {
			continue;
		} else {
			f_judge += 1;
			break;
		}
	}

	if(f_judge == 0) {
		for(i = latter; i < length; i++) {
			if(s[i-1] == s[length + latter - i - 1]) {
				continue;
			} else {
				l_judge += 1;
				break;
			}
		}
	}


	if(f_judge == 0 && l_judge == 0) {
		printf("Yes\n");
	} else {
		printf("No\n");
	}

	return 0;
}