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;

#define int long long

#define ii pair<int,int>
#define fi first
#define sc second
#define lwb lower_bound
#define upb upper_bound
#define all(x) (x).begin(),(x).end()

bool palin(string s) {
  int n = s.size();
  for (int i = 0; i < n / 2; i++) {
    if (s[i] != s[n - i - 1]) return 0;
  }
  return 1;
}

signed main() {
#ifdef _DEBUG
  freopen("in", "r", stdin);
#endif
  ios::sync_with_stdio(0); cin.tie(0);
  string s;
  cin >> s;
  int n = s.size();
  bool res = palin(s)
    && palin(s.substr(0, n / 2))
    && palin(s.substr(n / 2 + 1, n / 2));
  cout << (res ? "Yes" : "No") << '\n';
}