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<iostream>
#include<vector>
#include<algorithm>
#include<cstdio>
#include<string>
#include<stdio.h>
#include<stdlib.h>
#include<float.h>
#include<tuple>
#include<string.h>
#include<iomanip>
#include<stack>
#include<queue>

using namespace std;
#define ll long long
#define rep(i,n) for(ll i=0;i<n;i++)
#define ALLOF(c) (c).begin(), (c).end()
#define Pa pair<ll,ll>
const ll mod=1000000007;
const ll INF=10e12;
const ll inf=-1;
ll ABS(ll a){return max(a,-a);}


int main(void){
    string S;
    cin>>S;
    ll N=S.size();
    bool YES=true;
    rep(i,N) if(S.at(i)!=S.at(N-1-i)) YES=false;
    rep(i,(N-3)/2) if(S.at(i)!=S.at((N-3)/2-i)) YES=false;
    rep(i,(N-3)/2) if(S.at((N+1)/2+i)!=S.at(N-1-i)) YES=false;
    
    if(YES) cout<<"Yes"<<endl;
    else cout<<"No"<<endl;
    return 0;
    
}