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>
#define SORT(x) sort((x).begin(), (x).end())
#define RSORT(x) sort((x).rbegin(), (x).rend())
#define ALL(x) (x).begin(), (x).end()
#define rep(i, n) for (ll i = 0; i < n; i++)
#define reps(i, m, n) for (ll i = m; i < n; i++)
#define repr(i, m, n) for (ll i = m; i >= n; i--)
#define de(x) cout << #x << "=" << x << endl;
#define SP << " " <<
template <class T> bool maxi(T &a, const T &b) {
    if (a < b) { a = b; return 1;} return 0;}
template <class T> bool mini(T &a, const T &b) {
    if (b < a) { a = b; return 1;} return 0;}
#define dame {cout<< "-1" <<"\n"; return;}
#define INF2 1000000000000000037
#define INF  1000000007
#define MOD  1000000007
using namespace std;
using ll = long long;
using ld = long double;
using P  = pair<ll,ll>;
//--GLOBAL---------------------------------

//--MAIN-----------------------------------
void Main() {
    string S; cin>>S;
    ll N=S.size();
    bool ok=true;
    rep(i,N){
        if(S[i]!=S[N-1-i]) ok=false;
    }
    rep(i,(N-1)/2){
        if(S[i]!=S[(N-1)/2-1-i]) ok=false;
        if(S[N-1-i]!=S[(N+1)/2+i]) ok=false;
    }
    cout<< ( ok ? "Yes" : "No" )<<"\n";

}

//--START----------------------------------
int main() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(15);
    Main();
}
//-----------------------------------------