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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
using namespace std;

char s[233];

int main(){
	scanf("%s",s+1);
	int L = 1;
	int R = (strlen(s + 1) - 1) / 2;
	bool flag1 = true;
	for(int i = 1; L <= R; ){
		if(s[L] != s[R]){
			flag1 = false;
			break;
		}
		L += i;
		R -= i;
	}
	L = (strlen(s + 1) + 3) / 2;
	R = strlen(s + 1);
	bool flag2 = true;
	for(int i = 1; L <= R; ){
		if(s[L] != s[R]){
			flag2 = false;
			break;
		}
		L += i;
		R -= i;
	}
	bool flag3 = true;
	L = 1;
	R = strlen(s + 1);
	for(int i = 1; L <= R; ){
		if(s[L] != s[R]){
			flag2 = false;
			break;
		}
		L += i;
		R -= i;
	}
	if(flag1 && flag2) printf("Yes\n");
	else printf("No\n");
	return 0;
}