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: Wrong Answer)

#include <stdio.h>
int main(void){
 int N, i, j, k;
 char S[100] = {};
 char S_inv[100] = {};
 char A[100] = {};
 char A_inv[100] = {};
 char B[100] = {};
 char B_inv[100] = {};
 scanf("%s", S);
 N = 0;
 while(S[N] != 0){
 N++;
 } 
 for(i = 0; i <= N - 1; i++){
  S_inv[i] = S[N - 1 - i];
 }
 for(i = 0; i <= (N - 1) / 2 - 1; i++){
  A_inv[i] = S[(N - 1) / 2 - 1 - i];
 }
 for(i = 0; i <= (N - 1) / 2 - 1; i++){
  B_inv[i] = S[N - 1 - i];
 }
 for(i = 0; i <= (N - 1) / 2 - 1; i++){
   k = 0;
   if(S_inv[i] != S[i]){
     printf("No");
     break;
   }else if(A_inv[i] != S[i] && B_inv[i] != S[N - 1 - i]){
     printf("No");
     break;
   }else{
     k = 10;
   }
 }
 if(k == 10){
   printf("Yes");
 }
 return 0;
}