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:
/*Author: Satyajeet Singh, Delhi Technological University*/
    import java.io.*;
    import java.util.*;
    import java.text.*; 
    import java.lang.*;
    import java.math.*;

public class Main{
/*********************************************Constants******************************************/
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));        
    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static long mod=(long)1e9+7;
    static long mod1=998244353;
    static boolean sieve[];
    static ArrayList<Integer> primes;
    static ArrayList<Long> factorial;
    static HashSet<Pair> graph[];
/****************************************Solutions Begins***************************************/
    public static void main (String[] args) throws Exception {
        String st[]=br.readLine().split(" ");
        int x=Integer.parseInt(st[0]);
        int prev=0;
        ArrayList<Integer> list=new ArrayList<>();
        HashSet<Integer> valid=new HashSet<>();
        for(int i=1;i<=31;i++){
            valid.add((1<<i)-1);
        }
        int steps=0;
        while(true){
            if(x==0){
                list.add(1);
                steps++;
                break;
            }
            if(valid.contains(x)){
                break;
            }
            int xor=0;
            for(int i=31;i>=0;i--){
               
                if((x&(1<<i))!=0){
                    
                  //  debug(i);
                    x=x^(((1<<(i+1))-1));
                    list.add(i+1);
                    break;
                }
            }
          //  debug(x);
            
           steps++;
            if(valid.contains(x)){
                break;
            }
            x++;
            steps++;
           // debug(x);
           // break;
        }
        out.println(steps);
        for(int u:list){
            out.print(u+" ");
        }
/****************************************Solutions Ends**************************************************/
        out.flush();
        out.close();
    }

/****************************************Template Begins************************************************/
/***************************************Precision Printing**********************************************/
    static void printPrecision(double d){
        DecimalFormat ft = new DecimalFormat("0.00000000000000000"); 
        out.println(ft.format(d));
    }
/******************************************Graph*********************************************************/
    static void Makegraph(int n){
        graph=new HashSet[n];
        for(int i=0;i<n;i++){
            graph[i]=new HashSet<>();
        }
    }
    static void addEdge(int a,int b,int c){
        graph[a].add(new Pair(b,c));
    }    
/*********************************************PAIR********************************************************/
    static class PairComp implements Comparator<Pair>{
        public int compare(Pair p1,Pair p2){
            return p2.v-p1.v;
        }
    }
    static class Pair implements Comparable<Pair> {
        int u;
        int v;
        int index=-1;
        public Pair(int u, int v) {
            this.u = u;
            this.v = v;
        }
 
        public int hashCode() {
            int hu = (int) (u ^ (u >>> 32));
            int hv = (int) (v ^ (v >>> 32));
            return 31 * hu + hv;
        }
 
        public boolean equals(Object o) {
            Pair other = (Pair) o;
            return u == other.u && v == other.v;
        }
 
        public int compareTo(Pair other) {
            if(index!=other.index)
                return Long.compare(index, other.index);
            return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
        }
 
        public String toString() {
            return "[u=" + u + ", v=" + v + "]";
        }
    }
/******************************************Long Pair*************************************************/
    static class PairCompL implements Comparator<Pairl>{
        public int compare(Pairl p1,Pairl p2){
            if(p1.v>p2.v){
                return -1;
            }
            else if(p1.v<p2.v){
                return 1;
            }
            else{
                return 0;
            }
        }
    }
    static class Pairl implements Comparable<Pair> {
            long u;
            long v;
            int index=-1;
            public Pairl(long u, long v) {
                this.u = u;
                this.v = v;
            }
    
            public int hashCode() {
                int hu = (int) (u ^ (u >>> 32));
                int hv = (int) (v ^ (v >>> 32));
                return 31 * hu + hv;
            }
    
            public boolean equals(Object o) {
                Pair other = (Pair) o;
                return u == other.u && v == other.v;
            }
    
            public int compareTo(Pair other) {
                if(index!=other.index)
                    return Long.compare(index, other.index);
                return Long.compare(v, other.v)!=0?Long.compare(v, other.v):Long.compare(u, other.u);
            }
    
            public String toString() {
                return "[u=" + u + ", v=" + v + "]";
            }
        }
/*****************************************DEBUG***********************************************************/
    public static void debug(Object... o) {
        System.out.println(Arrays.deepToString(o));
    }
/************************************MODULAR EXPONENTIATION***********************************************/
    static long modulo(long a,long b,long c) {
        long x=1;
        long y=a;
        while(b > 0){
            if(b%2 == 1){
                x=(x*y)%c;
            }
            y = (y*y)%c; // squaring the base
            b /= 2;
        }
        return  x%c;
    }
/********************************************GCD**********************************************************/
    static long gcd(long x, long y)
    {
        if(x==0)
            return y;
        if(y==0)
            return x;
        long r=0, a, b;
        a = (x > y) ? x : y; // a is greater number
        b = (x < y) ? x : y; // b is smaller number
        r = b;
        while(a % b != 0)
        {
            r = a % b;
            a = b;
            b = r;
        }
        return r;
    }
/******************************************SIEVE**********************************************************/
    static void sieveMake(int n){
        sieve=new boolean[n];
        Arrays.fill(sieve,true);
        sieve[0]=false;
        sieve[1]=false;
        for(int i=2;i*i<n;i++){
            if(sieve[i]){
                for(int j=i*i;j<n;j+=i){
                    sieve[j]=false;
                }
            }
        }
        primes=new ArrayList<Integer>();
        for(int i=0;i<n;i++){
            if(sieve[i]){
                primes.add(i);
            }
        }        
    }
/********************************************End***********************************************************/
}