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 <stdlib.h>
#include <string.h>

#define rep(i,l,n) for(int (i)=(l);(i)<(n);(i)++)
#define min(a,b) (a)<(b)?(a):(b)
#define max(a,b) (a)>(b)?(a):(b)

int kaibun(char*,int,int);

int main(){
	char s[100]; scanf("%s",s);
	int N=strlen(s);
	if( kaibun(s,0,N-1) && kaibun(s,0,(N-1)/2-1) && kaibun(s,(N+3)/2-1,N-1) ){
		printf("Yes");
	}else{
		printf("No");
	}
	return 0;
}

int kaibun(char *s,int a,int b){
	while(a<b){
		if( *(s+a)==*(s+b) ){
			a++; b--;
		}else{
			return 0;
		}
	}
	return 1;
}