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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define INF INT_MAX
#define ll  long long
#define ull unsigned long long
#define rep(i,n) for(ll i=0; i<n; ++i)
#define FOR(i, s, e) for(ll i=s; i<e; ++i)
#define MOD 1000000007
using namespace std;
template<class T> inline bool chmin(T& a, T b) { if(a>b) {a=b; return true;} return false;}
template<class T> inline bool chmax(T& a, T b) { if(a<b) {a=b; return true;} return false;}

bool f(string S) {
  for(int i=0; i<S.size()/2; ++i) {
    if(S[i] != S[S.size()-1-i]) return false;
  }
  return true;
}

int main() {
  string S; cin >> S;
  int N = S.size();
  string S_a = S.substr(0, (N-1)/2);
  string S_b = S.substr((N+3)/2-1, N-((N+3)/2-1));
  if(f(S) && f(S_a) && f(S_b)) cout << "Yes" << endl;
  else cout << "No" << endl;
}