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 <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i, l, r) for (int i = (int)(l); i < (int)(r); i++)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((int)x.size())
template <class T> bool chmax(T &a, T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T> bool chmin(T &a, T b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}

template <class T> using V = vector<T>;
using P = pair<int, int>;

/*
 */

signed main() {
  string s;
  cin >> s;
  int n = sz(s);

  string t = s;
  reverse(all(t));

  string a = s.substr(0, n/2);
  string b = a;
  reverse(all(b));

  string x = s.substr(0, 3);
  string y = x;
  reverse(all(y));

  if(s == t and a == b and x == y){
    cout << "Yes" << endl;
  }else{
    cout << "No" << endl;
  }

  return 0;
}