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;
using ll = long long;

int main() {
    string S;cin >> S;
    int N;N = S.size();
    int N2 = (N - 1) / 2;
    string A = S.substr(0,N2);
    string B = S.substr(N2 + 1,N2);
    int flag = 0;
    if (N2 % 2 == 0){
        for (int i = 0;i < (N2 / 2);i++){
            if (A[i] != A[N2 - 1 - i]) flag = 1;
            if (B[i] != B[N2 - 1 - i]) flag = 1;
        }
    }
    else {
        for (int i = 0;i < ((N2 - 1) / 2);i++){
            if (A[i] != A[N2 - 1 - i]) flag = 1;
            if (B[i] != B[N2 - 1 - i]) flag = 1;
        }
    }
    for (int i = 0;i < N2;i++){
        if (S[i] != S[N - 1 - i]) flag = 1;
    }
    if (flag == 0) cout << "Yes" << endl;
    else cout << "No" << endl;
}