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: Wrong Answer)

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

#define lli long long int
#define st first
#define nd second
#define mp make_pair
#define pb push_back
#define INF 1000000007
#define endl "\n"

lli n,m;
string a,a2,a3;

int main(){
	cin>>a;
	for(int i=0;i<a.size();i++){
		if(a[i]!=a[a.size()-i-1]) {
			cout<<"NO";
			return 0;
		}
	}
	for(int i=0;i<(a.size()-1)/2;i++){
		a2.pb(a[i]);
	}
	for(int i=(a.size()+3)/2-1;i<a.size();i++){
		a3.pb(a[i]);
	}
	for(int i=0;i<a2.size();i++){
		if(a2[i]!=a2[a2.size()-i-1]) {
			cout<<"NO";
			return 0;
		}
	}
	for(int i=0;i<a3.size();i++){
		if(a3[i]!=a3[a3.size()-i-1]) {
			cout<<"NO";
			return 0;
		}
	}
	cout<<"YES";
}