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 Rust (1.42.0) (judge verdict: Runtime Error)

use proconio::input;

fn is_palindrome(s: String) -> bool {
  let r: String = s.chars().rev().collect();
  return s == r;
}
fn is_string_palindrome(s: String) -> bool {
  let n = s.len();
  is_palindrome((&s[0..n]).to_string()) && 
  is_palindrome((&s[..(n-1)/2]).to_string()) &&
  is_palindrome((&s[(n+3)/2+1..n]).to_string())
}

fn main() {
  input! {
    s: String,
  }
  if is_string_palindrome(s) {
    println!("Yes");
  } else {
    println!("No");
  }
}