Problem:
Score : 1000 points

Problem Statement

We will call a string x good if it satisfies the following condition:

Condition: x can be represented as a concatenation of two copies of another string y of length at least 1.

For example, aa and bubobubo are good; an empty string, a, abcabcabc and abba are not good.

Eagle and Owl created a puzzle on good strings.
Find one string s that satisfies the following conditions. It can be proved that such a string always exists under the constraints in this problem.

1 ≤ |s| ≤ 200

Each character of s is one of the 100 characters represented by the integers 1 through 100.

Among the 2^{|s|} subsequences of s, exactly N are good strings.

Constraints

1 ≤ N ≤ 10^{12}

Input

Input is given from Standard Input in the following format:

N

Output

In the first line, print |s|, the length of s.
In the second line, print the elements in s in order, with spaces in between. Any string that satisfies the above conditions will be accepted.

Sample Input 1

7

Sample Output 1

4
1 1 1 1

There are two good strings that appear as subsequences of s: (1,1) and (1,1,1,1). There are six occurrences of (1,1) and one occurrence of (1,1,1,1), for a total of seven.

Sample Input 2

299

Sample Output 2

23
32 11 11 73 45 8 11 83 83 8 45 32 32 10 100 73 32 83 45 73 32 11 10

Solution in Rust (1.15.1) (judge verdict: Accepted)

fn run() {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    let n: usize = s.trim().parse().unwrap();
    let mut k = 1;
    while (1 << (k + 1)) - 1 <= n {
        k += 1;
    }
    let mut a = vec![];
    let mut b = vec![];
    for i in 1..(k + 1) {
        a.push(i);
        b.push(i);
    }
    let mut h = k + 1;
    let mut t = n - ((1 << k) - 1);
    while t > 0 {
        let mut x = 0;
        while (1 << (x + 1)) <= t {
            x += 1;
        }
        t -= 1 << x;
        a.push(h);
        b.push(h);
        for i in (x..(b.len() - 1)).rev() {
            b[i + 1] = b[i];
        }
        b[x] = h;
        h += 1;
    }
    a.append(&mut b);
    let mut out = String::new();
    out.push_str(&format!("{}\n", a.len()));
    for a in a {
        out.push_str(&format!("{} ", a));
    }
    out.pop();
    println!("{}", out);
}

fn main() {
    run();
}