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)

fn main() {
    let s = read::<String>();
    let v: Vec<char> = s.chars().into_iter().collect();
    let n = v.len();

    // 回文かどうか
    for i in 0..n {
        if v[i] != v[n-1-i] {
            println!("No");
            return;
        }
    }
    for i in 0..(n-1)/2 {
        if v[i] != v[(n-1)/2-1-i] {
            println!("No");
            return;
        }
    }
    for i in (n-1)/2..n {
        if v[i] != v[n-1-i] {
            println!("No");
            return;
        }
    }
    println!("Yes");
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().parse().ok().unwrap()
}