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)

#define _CRT_SECURE_NO_WARNINGS
#pragma comment(linker, "/stack:512000000")
#include<utility>
#include<iostream>
#include<fstream>
#include<vector>
#include<array>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<functional>
#include<random>
#include<climits>
#include<queue>
#include<stack>
#include<deque>
#include<bitset>
#include<map>
#include<set>
#include<unordered_map>
#include<unordered_set>
#include<cassert>
#include<ccomplex>

#ifdef ONLINE_JUDGE
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
#pragma GCC optimize("unroll-loops")
#endif

//4Head

#define all(x) x.begin(), x.end()
#define mp make_pair
#define X first
#define Y second

using namespace std;

using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937_64 gen(time(nullptr));

ll const mod = 1e9 + 7;

namespace {
	ll mul(ll a, ll b) {
		ll val = a * b - (ll)((ld)a * b / mod) * mod;
		if (val < 0) val += mod;
		if (val >= mod) val -= mod;
		return val;
	}

	ll poww(ll a, ll b) {
		ll val = 1;
		a %= mod;
		while (b > 0) {
			if (b % 2) val = mul(a, val);
			a = mul(a, a);
			b >>= 1;
		}
		return val % mod;
	}

	ll inv(ll a) {
		return poww(a, mod - 2);
	}
}

int const maxn = 3e5 + 5;

bool check(string s) {
	int n = s.size();
	for (int i = 0; i < n; i++) {
		if (s[i] != s[n - 1 - i]) return 0;
	}
	return 1;
}

signed main() {
	ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	string s; cin >> s;
	int n = s.size();
	string a = "";
	for (int i = 0; i < (n - 1) / 2; i++) {
		a += s[i];
	}
	string b = "";
	for (int i = (n + 3) / 2 - 1; i < n; i++) {
		b += s[i];
	}

	if (check(a) and check(b)) {
		cout << "Yes";
	} else cout << "No";


	return 0;
}

/*
5
2 6 15 5 10

4
2 6 15 5

*/