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 (1.15.1) (judge verdict: Wrong Answer)

fn main() {
    println!("{:?}", solve(input()));
}

fn input() -> Vec<u8> {
    let mut buf = String::new();
    std::io::stdin().read_line(&mut buf).unwrap();
    buf.pop();
    buf.into_bytes()
}

fn solve(s: Vec<u8>) -> String {
    let n = s.len();
    if !is_plaindrome(&s) { return String::from("No"); }
    if !is_plaindrome(&s[..(n-1)/2].to_vec()) { return String::from("No"); }
    if !is_plaindrome(&s[((n+3)/2)-1..].to_vec()) { return String::from("No"); }
    String::from("Yes")
}

fn is_plaindrome(s: &Vec<u8>) -> bool {
    let l = s.len();
    let mut first: Vec<u8> = s.to_vec();
    let latter: Vec<u8> = first.split_off(l/2);
    let mut latter_iter = latter.iter();
    if s.len() % 2 != 0 {
        latter_iter.next().unwrap();
    }
    first.reverse();
    for (a, b) in first.iter().zip(latter_iter) {
        if a != b {
            return false;
        }
    }
    true
}