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)

//人正是因为度过了一段阴暗岁月，才会更渴求黎明的曙光
//青春不能回头。所以青春也没有终点。
#include <iostream>
#include <sstream>
#include <fstream>
#include <string>
#include <memory.h>
#include <vector>
#include <deque>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <algorithm>
#include <functional>
#include <utility>
#include <bitset>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstdio>
using namespace std;

#define REP(i,n) for(int (i)=0; i<(int)(n); i++)

const int MAXN = 0;
bool run(string s) {
	string t = s;
	reverse(t.begin(), t.end());
	if(t == s) return true;
	else return false;
}

int main(void) {
	string s,tmp1,tmp2;
	cin >> s;
	int N = s.size();
	s = "*"+s;
	tmp1 = s.substr(1, (N-1)/2);
	tmp2 = s.substr((N+3)/2, N-(N+3)/2+1);
	if(run(tmp1) && run(tmp2)) cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}