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>
using namespace std;
#define rep(i,n) rep2(i,0,n)
#define rep2(i,m,n) for(int i=m;i<(n);i++)
#define all(a) (a).begin(), (a).end()
typedef long long ll;
#define INF 1e9
typedef pair<int, int> P;

/* 　"assert"    assert(A：条件式)で　Aが真なら処理は続行、偽ならプログラムを終了する　*/

/*　　ラムダ式　　 auto a = [キャプチャ](パラメータ){ 処理 }  */

/* string型　メンバ関数　substr()    replace()  */

bool isOK(string s){
    //cout << "文字列　＝"<< s << endl;
    string prevS = s;
    reverse(s.begin(), s.end());
    string revS = s;
    if(revS == prevS){
        return true;
    }
    return false;
}

int main(){
 
    string s; cin >> s;
    auto n = s.size();
    string before = s.substr(0,(n-1)/2);
    string after = s.substr((n+3)/2-1, n);
 
    if (isOK(s)) {
        if (isOK(before)) {
            if (isOK(after)) {
                cout << "Yes" << endl;
                return 0;
            }
        }
    }
    cout << "No" << endl;
     
}