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;

fn is_palindrome(s: &str) -> bool {
// sが回文であるかどうかを判定する
	let mut flag = true;
	let v:Vec<char> = s.chars().collect();
	let L = v.len();
	for n in 0..(L/2) {
		if v[n] != v[L - 1 - n] {
			flag = false;
			break;
		};
	}
	flag
}

fn main() {
	let mut buf = String::new();
	io::stdin().read_line(&mut buf).expect("");
	let s = buf.trim().to_string();
	let mut flag = true;
	if !is_palindrome(&s) {
		flag = false;
	}
	if !is_palindrome(&s[..((s.len()-1)/2)]) {
		flag = false;
	}
	if !is_palindrome(&s[((s.len()+1)/2)..]) {
		flag = false;
	}
	if flag {
		println!("Yes");
	}else{
		println!("No");
	}
}