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: Wrong Answer)

#include <bits/stdc++.h>
#include <algorithm>
using namespace std;
int main() {
    string str;
    int len=0;
    bool flg=false;
    cin>>str;
    //回文
    string tstr=str;
    std::reverse(str.begin(), str.end());
    if(str==tstr)flg=true;
    else flg =false;
  	
    str=tstr;
    
    string fstr;
    string astr;
    //1~
    len=(str.length()-1)/2;
    
    fstr=str.substr(len+1);
    
    string tfstr=fstr;
    std::reverse(fstr.begin(), fstr.end());
    if(fstr==tfstr)flg=true;
    else flg =false;
    
    astr=str.substr((str.length()+3)/2-1,str.length());
	string tastr=astr;
    std::reverse(astr.begin(), astr.end());
    if(astr==tastr)flg=true;
    else flg =false;
  
    if(flg){
 	  cout<<"Yes"<<endl;
    }
    else{
      cout<<"No"<<endl;
    }
    //cout<<fstr<<endl;
    //cout<<astr<<endl;
    return 0;
}