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)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS

#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
#include <cstring>
#include <cstdlib>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <bitset>
#include <utility>
#include <numeric>
#include <iterator>

using namespace std;

using ll = long long;
using ull = unsigned long long;

const ll inf = 1ll << 60;
const ll mod = (ll)1e9 + 7;

#define all(v) v.begin(), v.end()
#define rall(v) v.rbegin(), v.rend()

#define print(s) cout << s;
#define println(s) cout << s << endl;
#define printd(s, f) cout << fixed << setprecision(f) << s << endl;


int main()
{
	string s;
	cin >> s;

	bool res = true;

	string t = s;
	reverse(all(s));
	if (s != t)
	{
		res = false;
	}

	string p = s.substr(0, s.length() / 2);
	string p0 = p;
	reverse(all(p0));
	if (p != p0)
	{
		res = false;
	}

	ll x = (s.length() + 3) / 2 - 1;
	string q = s.substr(x, s.length() - x);
	string q0 = q;
	reverse(all(q0));
	if (q != q0)
	{
		res = false;
	}

	println((res ? "Yes" : "No"));
}