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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
//N = number of atoms

int main()
{
  int i,j,k,l;
  char S[256];
  scanf("%s",S);

  int N = strlen(S);
  
  k = 0;
  for(i=0;i<(N/2);i++){
    j = N-1-i;
    if( S[i] == S[j] ){
      k++;
    }
  }
  if( k == N/2 ){
    k = 0;
    for(i=0;i<((N)/4);i++){
      j = (N-1)/2-1-i;
      if( S[i] == S[j] ){
        k++;
      }
    }
    if( k == N/4 ){
      k = 0;
      l = -1;
      for(i=(((N+3)/2)-1);i<(((N+3)/2)-1)+N/4;i++){
        l++;
        j = N-1-l;       
        if( S[i] == S[j] ){
          k++;
        }
      }
      if( k == N/4 ){
        printf("Yes\n");
      }
      else{
        printf("No\n");
      }
    }
    else{
      printf("No\n");
    }
  }
  else{
    printf("No\n");
  }
  
  
  return 0; 
  
}