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)

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

int main(){
    string s;
    cin >> s;
    vector<char> S(s.size());
    for(int i=0; i<s.size(); i++){
        S.at(i) = s.at(i);
    }
    int N = s.size();

    vector<char> F((N-1)/2);
    vector<char> FR((N-1)/2);
    vector<char> L((N-1)/2);
    vector<char> LR((N-1)/2);
    
    for(int i=0; i< (N-1)/2; i++){
        F.at(i)=S.at(i);
        FR.at(i)=S.at(i);
    }
    for(int i=(N+1)/2; i<N; i++){
        L.at(i- (N+1)/2)=S.at(i);
        LR.at(i- (N+1)/2)=S.at(i);
    }
    reverse(FR.begin(), FR.end());
    reverse(LR.begin(), LR.end());

    bool judge=true;

    for(int i=0; i< F.size(); i++){
        if(F.at(i) != FR.at(i)) judge = false;
    }
    for(int i=0; i < L.size(); i++){
        if(L.at(i) != LR.at(i)) judge = false;
    }
    
    if(judge){
        cout << "Yes";
    }else{
        cout << "No";
    }
}