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;

using ll = long long;
#define REP(i,n) for(int i=0; i < (n); ++i)
#define REPR(i,n) for(int i=(n); i >= 0; --i)
#define FOR(i, m, n) for(int i = (m); i < (n); ++i)

#define INF 1e9

bool kaibun(string S){
  int n = S.size();
  REP(i, n/2){
    if(S[i] != S[n-1-i]){
      // cout << "false" << endl;
      return false;
    }
  }
  // cout << "true" << endl;
  return true;
}
int main(){
  string S;
  cin >> S;
  int n = S.size();
  if(kaibun(S) && kaibun(S.substr(0, (n-1)/2))&& kaibun(S.substr((n+3)/2-1, n-(n+3)/2+1))){
    cout << "Yes" << endl;
  }
  else{
    cout << "No" << endl;
  }
  return 0;
}