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: Accepted)

use proconio::input;

fn main() {
    input! {
        s: String,
    }

    solve(s);
}

fn solve(s: String) {
    if !is_palindrome(&s) {
        println!("No");
        return;
    }
    let mut begin = 0;
    let mut end = (s.len() - 1) / 2;
    if !is_palindrome(&s[begin..end].to_string()) {
        println!("No");
        return;
    }
    begin = (s.len() + 3) / 2 - 1;
    end = s.len();
    if !is_palindrome(&s[begin..end].to_string()) {
        println!("No");
        return;
    }

    println!("Yes");
}

fn is_palindrome(s: &String) -> bool {
    s.eq(&s.chars().rev().collect::<String>())
}