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++ (GCC 9.2.1) (judge verdict: Accepted)

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

#define f first
#define s second
#define ins insert
#define eb emplace_back
#define endl '\n'
#define ll long long
#define ld long double
#define vl vector<ll>
#define tc ll t; cin >> t; while (t--)2 2
#define yes cout << "Yes" << endl
#define no cout << "No" << endl
#define sz(c) (ll) c.size()
#define all(c) c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define trav(x,c) for(auto &x : c)
#define _for(i,a,n) for(ll i = a ; i < n ; ++i)
#define _rfor(i,a,n) for(ll i = n-1 ; i >= a ; --i)
#define dx(x) cout << #x << ": " << x << endl;
#define dc(c) trav(x,c) cout << x << " "; cout << endl;
#define godspeed ios::sync_with_stdio(0); cin.tie(0);

bool is_palin(string s) {
	string t = s;
	reverse(all(t));
	return (s == t) ? 1 : 0;
}

string palin(string s) {
	string t;
	ll n = sz(s);
	t = s;
	if(!is_palin(t)) return "No";

	t = s.substr(0,(n-1)/2);
	if(!is_palin(t)) return "No";

	t = s.substr((n+3)/2,n - 1 - ((n+3)/2));
	if(!is_palin(t)) return "No";

	return "Yes";

}

int main() {
    godspeed;
 	string s; cin >> s;
 	cout << palin(s);
}