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 solve(){
    let mut s = String::new();
    io::stdin().read_line(&mut s).unwrap();
    s = s.trim().to_owned();
    let n = s.len();
    let mut f = true;
    let s = s.as_bytes();
    for i in 0..n{
        if s[i] != s[n-1-i]{
            f = false;
            break;
        }
    }
    for i in 0..(n/2){
        if s[i] != s[n/2-1-i]{
            f = false;
            break;
        }
        if s[n-1-i] != s[n/2+1+i]{
            f = false;
            break;
        }
    }
    if f{println!("Yes")}
    else{println!("No")}
}


fn main(){
    let mut t_cases = 1;
    // let mut buf = String::new();
    // io::stdin().read_line(&mut buf);
    // t_cases = buf.trim().parse::<i32>();
    for _ in 0..t_cases{
        solve();
    }
}