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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using Graph = vector<vector<int>>;
typedef long long ll;
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define EFOR(i,a,b) for(int i=(a);i<=(b);++i)
#define REP(i, n) FOR(i,0,n)
#define INF 1e12
const int MOD = 1000000007;
typedef pair<ll,ll> P;

bool Palindrome(string s, int n){
    int f =  0;
    bool res = true;
    while(f != n){
        if(s[f] != s[n]){
            res = false;
            break;
        }
        f++;
        n--;
    }
    return res;
}

int main(void){
    string s;
    int n,n1,n2;
    cin >> s;
    n = s.length() - 1;
    n1 = (n - 1) / 2;
    n2 = (n + 3) / 2;
    if(     Palindrome(s,n) 
        &&  Palindrome(s.substr(0, n1+1), n1)
        &&  Palindrome(s.substr(n2, n), n - n2)) cout << "Yes" << endl;
    else cout << "No" << endl;
    return 0;
}