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 s;
  cin >> s;
  int N = s.size();
  bool A = false;
  
  for(int i = 0; i <= (N - 3) / 4 ; i++){
    if(s.at(i) == s.at((N - 1) / 2 - 1 - i)){
    }
    else{
      A = true;
      break;
    }
  }
  
  if(A == true){
    cout << "No" << endl;
    return 0;
  }
  
  A = false;
  
  for(int i = 0; i <= (N - 1) / 4 ; i++){
    if(s.at((N + 3) / 2 - 1 + i) == s.at(N - 1 - i)){
    }
    else{
      A = true;
      break;
    }
  }
  
  if(A == true){
    cout << "No" << endl;
    return 0;
  }
  
  A = false;
  
  for(int i = 0; i <= (N - 1) / 2 ; i++){
    if(s.at(i) == s.at(N - 1 - i)){
    }
    else{
      A = true;
      break;
    }
  }
  
  if(A == true){
    cout << "No" << endl;
    return 0;
  }
  
  cout << "Yes" << endl;
  
}