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>
#include <algorithm>
using namespace std;
#define rep(i,n) for (ll i = 0; i < (n); i++)
#define repd(i,a,b) for (ll i = (a); i < (b); i++)
#define INF LLONG_MAX
#define mod 1000000007
typedef long long ll;
int judge(string a, int l)
{
	rep(i, l/2)
	{
		if(a[i] != a[l-i-1]) return false;
	}
	return true;
}
int main(void)
{
	string S;
	cin >> S;
	int L = S.size();
	if(!judge(S, L))
	{
		cout << "No" << endl;
		return 0;
	}
	string H = S.substr(0, (L-1)/2);
	if(!judge(H, H.size()))
	{
		cout << "No" << endl;
		return 0;
	}
	string T = S.substr((L+3)/2-1,L-(L+3)/2+1);
	if(!judge(T, T.size()))
	{
		cout << "No" << endl;
		return 0;
	}
	cout << "Yes" << endl;
	return 0;
}