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::stdin;

fn main() {
    let mut s = String::new();
    stdin().read_line(&mut s).ok();

    println!("{}", if solve(&s) { "Yes" } else { "No" });
}

fn solve(s: &str) -> bool {
    let n = s.len();
    if !is_p(s) {
        return false;
    }

    let s1 = &s[..((&s.len() - 1) / 2)];
    if !is_p(s1) {
        return false;
    }

    let s2 = &s[(n + 3) / 2 - 1..];
    if !is_p(s2) {
        return false;
    }

    true
}

fn is_p(s: &str) -> bool {
    let v: Vec<char> = s.trim().chars().collect();

    for i in 0..v.len() / 2 {
        if v[i] != v[v.len() - i - 1] {
            return false;
        }
    }

    true
}