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 ll long long
#define mp make_pair
#define pb push_back
#define pii pair<int,int>
#define pll pair<ll,ll>
#define x first
#define y second
#define mod 1e9+7
#define f(i,a,b) for(ll i=a;i<b;i++)
#define rof(i,a,b) for(ll i=a;i>b;i--)

ll lcm(ll a, ll b){
    return a / __gcd(a, b) * b;
}
//memset can initialize to 0 or -1 ONLY
// memset(a,0,sizeof(a[0][0])*n*n); //2D array
//memset(a, 0, sizeof(a));  //1D array
//memset(str, 't', sizeof(str)); // string
void solve()
{
  string s;cin>>s;
  
  bool palin =true,palin1=true,palin2=true;
  int st=0,e=s.length()-1;
  int n=s.length();
  while(st!=e)
  {
      if(s[st]!=s[e])
      {
          palin=false;
      
          break;
      }
      st++;e--;
      
  }
  
  st=0;e=(n-1)/2 -1;
    while(st<e)
  {
      if(s[st]!=s[e])
      {
          palin1=false;
      
          break;
      }
      st++;e--;
      
  }
  
  st=(n+3)/2 -1;e=s.length()-1;
  
      while(st<e)
  {
      if(s[st]!=s[e])
      {
          palin2=false;
      
          break;
      }
      st++;e--;
      
  }
 // cout<<palin<<palin1<<palin2<<" ";
  if(palin&&palin1&&palin2)
  cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
}


int main()
{

ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    ll t=1;
   // cin>>t;
    while(t--)
    {
    
    solve();
    
    }

}