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 rep(i, n) for (int i = 0; i < (int)(n); i++)
#define srt(i) sort(i.begin(), i.end())
#define rvt(i) sort(i.begin(), i.end(), greater<int>())

int main(){
  string s;
  cin >> s;

  bool ans = true;
  int half=(s.size()-1)/2;

  // all
  for(int i=0; i<half; i++){
    if(s.at(i) != s.at(s.size()-i-1)) ans=false;
  }

  // left
  for(int i=0; i<half/2; i++){
    if(s.at(i) != s.at(half-i-1)) ans=false;
  }

  // right
  for(int i=half; i<(s.size()-1)/2; i++){
    if(s.at(i) != s.at(s.size()-i-1)) ans=false;
  }

  if(ans)  cout << "Yes" << endl;
  else cout << "No" << endl;
}