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)

// -*- coding:utf-8-unix -*-

use proconio::input;

// ABC086C - Traveling
// https://atcoder.jp/contests/abs/fasks/arc089_a

fn is_palindrome(s: &str) -> bool {
    // 長さ5のとき、
    // 0 1 2 3 4

    let end = if s.len() % 2 == 1 {
        (s.len() - 1) / 2
    } else {
        s.len() / 2
    };
    for i in 0..end {
        if s[i..i + 1] != s[(s.len() - 1 - i)..(s.len() - i)] {
            return false;
        }

    }
    return true;

}


fn is_strong_palindrome(s: &str) -> bool {
    if !is_palindrome(s) {
        return false;
    }


    let ins: &str = &s[..(s.len() - 1) / 2];
    if !is_palindrome(&ins) {
        return false;
    }


    let ins2: &str = &s[(s.len() + 3) / 2 - 1..s.len()];
    if !is_palindrome(&ins2) {
        return false;
    }


    return true;
}

fn main() {
    input! {
        s: String,
    }

    let ans = is_strong_palindrome(&s[..]);
    let yes = "Yes";
    let no = "No";
    println!("{}", if ans { yes } else { no });
}