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>

#define N 100

int main(){
  char s[100];
  int n, noStrongKaibun=0;
  scanf("%s",&s);
  n = strlen(s);
//  printf("%d",n);
  
  for(int i=0; i<n-i-1 ;i++){
    if(s[i]!=s[n-i-1]){
      noStrongKaibun=1;
      break;
    }
  }
  
  for(int i=0; i<(n-3)/2-i ;i++){
    if(s[i]!=s[(n-3)/2-i]){
      noStrongKaibun=1;
      break;
    }
  }
  
  for(int i=0; (n+1)/2+i<n-i-1 ;i++){
    if(s[(n+1)/2+i]!=s[n-i-1]){
      noStrongKaibun=1;
      break;
    }
  }
  
  if(noStrongKaibun==1)
    printf("No");
  else
    printf("Yes");
  
  return 0;
}