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;

int main()
{
    string M;
    cin >> M;
    int N=M.size();
    bool A=true;
    for(int i=0; i<N; i++)
    {
        if(!(M.at(i)==M.at(N-1-i)))
        {
            A=false;
        }
        if((i==(N-1-i))||(i==N/2))
        {
            break;
        }
    }
        for(int i=0; i<N; i++)
    {
        if(!(M.at(i)==M.at(((N-1)/2)-1-i)))
        {
            A=false;
        }
        if(i==(((N-1)/2)-1-i)||(N/4))
        {
            break;
        }
    }

    for(int i=0; i<N; i++)
    {
        if(!(M.at(((N+3)/2)-1+i)==M.at((N-1)-i)))
        {
            A=false;
        }
        if(((((N+3)/2)-1+i)==((N-1)-i))||N/4)
        {
            break;
        }
    }

    if(A==true)
    {
        cout << "Yes" <<endl;
    }
    else
    {
        cout << "No" <<endl;
    }
    
}