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>

#define BUF 101

int judge(char *str, int size) {
  for (int i = 0; i < (size / 2); i++) {
    // printf("%c %c\n", str[i], str[size - 1 - i]);
    if (str[i] != str[size - 1 - i]) return 0;
  }

  return 1;
}

int main(void) {
  char str[BUF];
  memset(str, '\0', BUF);

  scanf("%s", str);
  int str_len = strlen(str);
  // printf("%d\n", str_len);
  if (judge(str, str_len)) {
    int str_half_len = (str_len - 1) / 2;
    if (judge(str, str_half_len) && judge((str + (str_len + 3) / 2 - 1), str_half_len)) {
      printf("Yes\n");
      return 0;
    }
  }
  printf("No\n");

  return 0;
}