Problem: 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.

Examples

Input

7


Output

4
1 1 1 1


Input

299


Output

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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	void go(ArrayDeque<Integer> q, long x) {
		if (x == 1) {
			return;
		}
		if (x % 2 == 0) {
			go(q, x / 2);
			q.addLast(q.size() + 1);
		} else {
			go(q, x - 1);
			q.addFirst(q.size() + 1);
		}
	}
	
	void solve() throws IOException {
		long x = nextLong() + 1;
		ArrayDeque<Integer> q = new ArrayDeque<>();
		go(q, x);
		
		List<Integer> ans = new ArrayList<>();
		for (int i = 1; i <= q.size(); i++) {
			ans.add(i);
		}
		
		ans.addAll(q);
		out.println(ans.size());
		for (int xx : ans) {
			out.print(xx + " ");
		}
		out.println();
	}

	Main() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new Main();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}