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: Accepted)

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("faild to read char") as char)
        .take_while(|c| !c.is_whitespace())
        .collect();
    token
        .parse()
        .ok()
        .expect("faild to parse token")
}

fn main(){
    let s: Vec<u8> = read::<String>().into_bytes();
    let half = s.len() / 2;
    for i in 0..s.len() {
        if s[i] != s[s.len() - 1 - i] {
            println!("No");
            return;
        }
        
        if i < half && s[i] != s[half - i - 1] {
            println!("No");
            return;
        }

        if i > half && s[i] != s[s.len() - i + half] {
            println!("No");
            return;
        }
    }
    println!("Yes");
}