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<iostream>
#include<string>
#include<algorithm>
#include<bits/stdc++.h>
#include<math.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n); i++)
#define rep1(i,n) for(int i = 1; i < (n); i++)

int main(){
    string s;
    cin >> s;
    int n = s.length();
    bool ok = true;
    string t;
    
    rep(i, (n-1)/2){
        t[i] = s[(n-3)/2-i];
    }
    rep(i, (n-1)/2){
        if(s[i] != t[i]){
            ok = false;
        }
        if(s[i] != s[n-i]){
            ok = false;
        }
    }rep(i, (n-1)/2){
        t[n-1-i] = s[(n+1)/2+i];
    }
    rep(i, (n-1)/2){
        if(s[n-1-i] != t[n-1-i]){
            ok = false;
        }
    }
    if(ok){
        cout << "Yes" << endl;
    }else {
        cout << "No" << endl;
    }
}