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)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char) 
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn isPalindrome(s: &str, n: usize) -> bool {
	let startStr: &str = &s[0..n/2];
    let endStr: &str = if n%2 == 0 { &s[n/2..n] } else { &s[n/2+1..n] };
    if startStr == endStr { return true } else { return false }
}

fn main() {
	let s: String = read();
    let n: usize = s.len();
    if !isPalindrome(&s, n) { println!("No"); return; };
    if !isPalindrome(&s[0..n/2], n/2) { println!("No"); return; };
    if !isPalindrome(if n%2 == 0 { &s[n/2..n] } else { &s[n/2+1..n] }, n/2) { println!("No"); return; };
    println!("Yes");
}