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>
//#pragma once
using namespace std;
#define repd(i,a,b) for (int i=(a);i<(b);i++)
#define rep(i,n) repd(i,0,n)
typedef long long ll;
typedef long double lb;
#define int long long
#define double long double
typedef pair<double,double> P;
const ll INF = 4*1e18;
const int MOD = 1000000007;
const double EPS = 1e-10;
const double PI = 3.141592653589793;
template<class T> inline bool chmin(T& a, T b){if(a>b){a = b;return 1;}return 0;}
template<class T> inline bool chmax(T& a, T b){if(a<b){a = b;return 1;}return 0;}
const pair<int,int> fd[] = {make_pair(1,0),make_pair(-1,0),make_pair(0,1),make_pair(0,-1)}; 

bool o(string s){
  int n = s.length();
  bool ret = true;
  rep(i,n/2){
    ret = ret && (s[i] == s[n-i-1]);
  }
  return ret;
}

signed main(){
  string s;
  cin>>s;
  cout << ((o(s) && o(s.substr(0,(s.length()-1)/2)) && o(s.substr((s.length()+3)/2-1, (s.length()-1)/2))) ? "Yes" : "No") << endl;
}