Problem: Cat Furrier Transform is a popular algorithm among cat programmers to create longcats. As one of the greatest cat programmers ever exist, Neko wants to utilize this algorithm to create the perfect longcat.

Assume that we have a cat with a number x. A perfect longcat is a cat with a number equal 2^m - 1 for some non-negative integer m. For example, the numbers 0, 1, 3, 7, 15 and so on are suitable for the perfect longcats.

In the Cat Furrier Transform, the following operations can be performed on x:

  * (Operation A): you select any non-negative integer n and replace x with x ⊕ (2^n - 1), with ⊕ being a [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  * (Operation B): replace x with x + 1. 



The first applied operation must be of type A, the second of type B, the third of type A again, and so on. Formally, if we number operations from one in the order they are executed, then odd-numbered operations must be of type A and the even-numbered operations must be of type B.

Neko wants to produce perfect longcats at industrial scale, thus for each cat Neko only wants to perform at most 40 operations. Can you help Neko writing a transformation plan?

Note that it is not required to minimize the number of operations. You just need to use no more than 40 operations.

Input

The only line contains a single integer x (1 ≤ x ≤ 10^6).

Output

The first line should contain a single integer t (0 ≤ t ≤ 40) — the number of operations to apply.

Then for each odd-numbered operation print the corresponding number n_i in it. That is, print ⌈ t/2 ⌉ integers n_i (0 ≤ n_i ≤ 30), denoting the replacement x with x ⊕ (2^{n_i} - 1) in the corresponding step.

If there are multiple possible answers, you can print any of them. It is possible to show, that there is at least one answer in the constraints of this problem.

Examples

Input


39


Output


4
5 3 

Input


1


Output


0


Input


7


Output


0

Note

In the first test, one of the transforms might be as follows: 39 → 56 → 57 → 62 → 63. Or more precisely:

  1. Pick n = 5. x is transformed into 39 ⊕ 31, or 56. 
  2. Increase x by 1, changing its value to 57. 
  3. Pick n = 3. x is transformed into 57 ⊕ 7, or 62. 
  4. Increase x by 1, changing its value to 63 = 2^6 - 1. 



In the second and third test, the number already satisfies the goal requirement.

Solution (in JAVA):

Solution:
// Change Of Plans BABY.... Change Of Plans //
import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class catFurierTransform {

    static void MainSolution() {
        n = ni();
        int c = 0;
        StringBuilder sb = new StringBuilder();
        while (!check(n)) {

            if ((c & 1) == 0) {
                for (int i = 21; i >= 0; i--) {
                    if (n >= (1 << i)) {
                        n ^= (1 << (i + 1)) - 1;
                        sb.append((i + 1) + " ");
                        break;
                    }
                }
            } else n++;

            c++;
        }
        pl(c);
        pl(sb);
    }

    static boolean check(int x) {
        for (int i = 23; i >= 0; i--) {
            if (x == (1 << i) - 1) return true;
        }
        return false;
    }

    /*----------------------------------------------------------------------------------------------------------------*/
    //THE DON'T CARE ZONE BEGINS HERE...\\
    static int mod9 = 1000000007;
    static int n, m, l, k, t, mod = 998244353;
    static AwesomeInput input = new AwesomeInput(System.in);
    static PrintWriter pw = new PrintWriter(System.out, true);

    static class AwesomeInput {
        private InputStream letsDoIT;
        private byte[] buf = new byte[1024];
        private int curChar;
        private int numChars;
        private SpaceCharFilter filter;

        private AwesomeInput(InputStream incoming) {
            this.letsDoIT = incoming;
        }

        public int read() {
            if (numChars == -1)
                throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = letsDoIT.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars <= 0)
                    return -1;
            }
            return buf[curChar++];
        }

        private long ForLong() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = read();
            }
            long res = 0;
            do {
                if (c < '0' || c > '9')
                    throw new InputMismatchException();
                res *= 10;
                res += c - '0';
                c = read();
            }
            while (!isSpaceChar(c));
            return res * sgn;
        }

        private String ForString() {
            int c = read();
            while (isSpaceChar(c))
                c = read();
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = read();
            }
            while (!isSpaceChar(c));

            return res.toString();
        }

        public boolean isSpaceChar(int c) {
            if (filter != null)
                return filter.isSpaceChar(c);
            return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
        }

        public interface SpaceCharFilter {
            boolean isSpaceChar(int ch);
        }
    }

    // functions to take input//
    static int ni() {
        return (int) input.ForLong();
    }

    static String ns() {
        return input.ForString();
    }

    static long nl() {
        return input.ForLong();
    }

    static double nd() throws IOException {
        return Double.parseDouble(new BufferedReader(new InputStreamReader(System.in)).readLine());
    }

    //functions to give output
    static void pl() {
        pw.println();
    }

    static void p(Object o) {
        pw.print(o + " ");
    }

    static void pws(Object o) {
        pw.print(o + "");
    }

    static void pl(Object o) {
        pw.println(o);
    }

    public static int[] fastSort(int[] f) {
        int n = f.length;
        int[] to = new int[n];
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] & 0xffff)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] & 0xffff]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        {
            int[] b = new int[65537];
            for (int i = 0; i < n; i++) b[1 + (f[i] >>> 16)]++;
            for (int i = 1; i <= 65536; i++) b[i] += b[i - 1];
            for (int i = 0; i < n; i++) to[b[f[i] >>> 16]++] = f[i];
            int[] d = f;
            f = to;
            to = d;
        }
        return f;
    }

    public static void main(String[] args) {      //threading has been used to increase the stack size.

        new Thread(null, null, "Vengeance", 1 << 25)  //the last parameter is stack size desired.
        {
            public void run() {
                try {
                    double s = System.currentTimeMillis();
                    MainSolution();
                    //pl(("\nExecution Time : " + ((double) System.currentTimeMillis() - s) / 1000) + " s");
                    pw.flush();
                    pw.close();
                } catch (Exception e) {
                    e.printStackTrace();
                    System.exit(1);
                }
            }
        }.start();
    }
}