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 <iostream>
using namespace std;
#include <string>

int main() {
  string s;
  int n;
  int i,j,c,k;
  string ans="Yes";
  
  cin >> s;

  n=(int)s.length();
  //条件1
  for(i=0,j=n-1;i<n/2;i++,j--) {
    if(s[i]!=s[j]) {
      ans="No";
    }
  }
  //条件２
  if(ans=="Yes") {
    for(i=0,j=((n-1)/2)-1;i<((n-1)/2)/2;i++,j--) {
      if(s[i]!=s[j]) {
      
      ans="No";
      }
    }
  }
  c=i+1;
  k=0;
  //条件3
  if(ans=="Yes") {
    for(i=((n+3)/2)-1,j=n-1;k<c;i++,j--,k++) {
      if(s[i]!=s[j]) {
        ans="No";
      }
    }
  }
  cout << ans << endl;




  return 0;
}