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 (1.42.0) (judge verdict: Accepted)

use proconio::input;

#[allow(unused_imports)]
use proconio::marker::{Bytes, Chars, Usize1};
#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::ops::*;

#[derive(Clone, Debug, Default)]
struct Struct;

//#[proconio::fastout]
fn main() {
    input! {
        s: Bytes,
    }

    let len = s.len();
    let a = (s.len() - 1) / 2;
    let b = (0..a).all(|i| s[i] == s[a - 1 - i] && s[i] == s[a + 1 + i] && s[i] == s[len - 1 - i]);

    println!("{}", if b { "Yes" } else { "No" });
}