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:
import java.io.*;
import java.io.ObjectInputStream.GetField;
import java.math.*;
import java.util.*;

/**
 *
 * @author Saju
 *
 */

public class Main {

	private static int dx[] = { -2, -3, -2, -1, -1, 1};
	private static int dy[] = { 1, -1, -1, -2, -3, -2};

	private static final long INF = (long) (1e15);
	private static final long NEG_INF = Long.MIN_VALUE;
	private static final double EPSILON = 1e-10;

	private static final int MAX = 1007;
	private static final int MOD = 10007;
	
	private static final int MAXN = 100007;
	private static final int MAXA = 10000009;
	private static final int MAXLOG = 22;

	public static void main(String[] args) {

		InputReader in = new InputReader(System.in);
		PrintWriter out = new PrintWriter(System.out);
		

/*

*/
		
		
		twoPowers[0] = 1;
		for(int i = 1; i < 63; i++) {
			twoPowers[i] = twoPowers[i - 1] * 2;
		}
		
		
	
		long x = in.nextLong();
		
		List<Integer> list = new ArrayList<Integer>();
		
		int step = 1;
		while(true) {
			int logx = log(x, 2);
//			System.out.println(x + " " + logx);
			if(x == twoPowers[logx + 1] - 1) {
				break;
			}
			if(step % 2 == 0) {
				x++;
			}
			else {
				if(twoPowers[logx] != x) {
					logx++;
				}
				x = x ^ (twoPowers[logx] - 1);
				list.add(logx);
			}
			
			step++;
		}
		step--;
		out.println(step);
		for(Integer e : list) {
			out.print(e + " ");
		}
		out.println();
		
		out.flush();
		out.close();
		System.exit(0);
	}

	static long twoPowers[] = new long[63];
	
	private static class Pair<T>{
		T a;
		T b;
		Pair(T a, T b){
			this.a = a;
			this.b = b;
		}
	}
	
	private static class SegmentTree{
		int n;
		private final int MAXN = 100007;
		private long[] tree = new long[MAXN << 2];
		private long[] lazy = new long[MAXN << 2];
		private long[] arr = new long[MAXN];
		
		
		/***
		 * 
		 * arr is 1 based index.
		 * 
		 * @param arr
		 * 
		 */
		SegmentTree(int n, long[] arr){
			this.n = n;
			for(int i = 0; i < arr.length; i++) {
				this.arr[i] = arr[i];
			}
		}
		
		void build(int index, int left, int right) {
			if(left == right) {
				tree[index] = arr[left];
			}
			else {
				int mid = (left + right) / 2;
				
				build(index * 2, left, mid);
				build((index * 2) + 1, mid + 1, right);
				tree[index] = max(tree[(index * 2)], tree[(index * 2) + 1]);
			}
		}
		
		long query(int node, int left, int right, int start, int end) {
			
			if(left > end || right < start) {
				return NEG_INF;
			}
			
			if(left >= start && right <= end) {
				return tree[node];
			}
			int mid = (left + right) / 2;
			long val1 = query(2 * node, left, mid, start, end);
			long val2 = query(2 * node + 1, mid + 1, right, start, end);
			
			return max(val1, val2);
		}
		
		void update(int node, int left, int right, int idx, long val) {
			if(left == right) {
				tree[node] += val;
			}
			else {
				int mid = (left + right) / 2;
				if(idx <= mid) {
					update(2 * node, left, mid, idx, val);
				}
				else {
					update(2 * node + 1, mid + 1, right, idx, val);
				}
				tree[node] = max(tree[(2 * node) + 1], tree[(2 * node)]);
			}
		}
		
		

		 void updateRange(int node, int start, int end, int l, int r, long val)
		{
		    if(lazy[node] != 0)
		    { 
		        // This node needs to be updated
		        tree[node] =  lazy[node];    // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];                  // Mark child as lazy
		            lazy[node*2+1] = lazy[node];                // Mark child as lazy
		        }
		        lazy[node] = 0;                                  // Reset it
		    }
		    if(start > end || start > r || end < l)              // Current segment is not within range [l, r]
		        return;
		    if(start >= l && end <= r)
		    {
		        // Segment is fully within range
		        tree[node] =  val;
		        if(start != end)
		        {
		            // Not leaf node
		            lazy[node*2] = val;
		            lazy[node*2+1] = val;
		        }
		        return;
		    }
		    int mid = (start + end) / 2;
		    updateRange(node*2, start, mid, l, r, val);        // Updating left child
		    updateRange(node*2 + 1, mid + 1, end, l, r, val);   // Updating right child
		    tree[node] = max(tree[node*2], tree[node*2+1]);        // Updating root with max value 
		}

		 long queryRange(int node, int start, int end, int l, int r)
		{
		    if(start > end || start > r || end < l)
		        return 0;         // Out of range
		    if(lazy[node] != 0)
		    {
		        // This node needs to be updated
		        tree[node] =  lazy[node];            // Update it
		        if(start != end)
		        {
		            lazy[node*2] = lazy[node];         // Mark child as lazy
		            lazy[node*2+1] = lazy[node];    // Mark child as lazy
		        }
		        lazy[node] = 0;                 // Reset it
		    }
		    if(start >= l && end <= r)             // Current segment is totally within range [l, r]
		        return tree[node];
		    int mid = (start + end) / 2;
		    long p1 = queryRange(node*2, start, mid, l, r);         // Query left child
		    long p2 = queryRange(node*2 + 1, mid + 1, end, l, r); // Query right child
		    return max(p1, p2);
		}
		
		 void buildRange(int node, int low, int high){
			if(low == high){
				tree[node] =  arr[low];
				return;
			}
			
			int mid = (low + high) / 2;
			int left = node << 1;
			int right = left | 1;
			buildRange(left, low, mid);
			buildRange(right, mid + 1, high);
			tree[node] = max(tree[left], tree[right]);
		}
		
		
		void printSegmentTree() {
			System.out.println(Arrays.toString(tree));
		}
		
	}
	
	private static class KMP{
		
		private static char[] t;
		private static char[] s;
		
		public int kmp(char[]t, char[]s) {
			this.t = t;
			this.s = s;
			return this.kmp();
		}
		
		private int kmp() {
			
			List<Integer> prefixTable = getPrefixTable(s);
			
			int match = 0;
			int i = 0;
			int j = 0;
			
			int n = t.length;
			int m = s.length;
			while(i < n) {
				if(t[i] == s[j]) {
					if(j == m - 1) {
						match++;
						j = prefixTable.get(j - 1);
						continue;
					}
					i++;
					j++;
				}
				else if(j > 0) {
					j = prefixTable.get(j - 1);
				}
				else {
					i++;
				}
			}
			
			return match;
		}

		/***
		1. We compute the prefix values π[i] in a loop by iterating 	<br/>
				from i=1 to i=n−1 (π[0] just gets assigned with 0).		<br/><br/>
		2. To calculate the current value π[i] we set the variable j 	<br/>
				denoting the length of the best suffix for i−1. Initially j=π[i−1].	<br/>
		3. Test if the suffix of length j+1 is also a prefix by		 <br/><br/>
				comparing s[j] and s[i]. If they are equal then we assign π[i]=j+1, 	<br/>
				otherwise we reduce j to π[j−1] and repeat this step.		<br/><br/>
		4. If we have reached the length j=0 and still don't have a match,	<br/> 
				then we assign π[i]=0 and go to the next index i+1.		<br/><br/>
				
		@param
	 	pattern(String)
	***/
		private List<Integer> getPrefixTable(char[] pattern){
			
			List<Integer> prefixTable = new ArrayList<Integer>();
			int n = pattern.length;
			for(int i = 0; i < n; i++) {
				prefixTable.add(0);
			}
			
			for(int i = 1; i < n; i++) {
				for(int j = prefixTable.get(i - 1); j >= 0;) {
					if(pattern[j] == pattern[i]) {
						prefixTable.set(i, j + 1);
						break;
					}
					else if(j > 0){
						j = prefixTable.get(j - 1);
					}
					else {
						break;
					}
				}
			}
			
			return prefixTable;
		}
	}
	


	private static class Point {
		long x;
		long y;
		Point(long x, long y){
			this.x = x;
			this.y = y;
		}
		
		@Override
		public boolean equals(Object obj) {
			Point ob = (Point) obj;
			if(this.x == ob.x && this.y == ob.y){
				return true;
			}
			return false;
		}
		@Override
		public String toString() {
			return this.x + " " + this.y;
		}
	}

	private static int pow(int base, int pow) {
		int val = 1;
		for(int i = 1; i <= pow; i++) {
			val *= base;
		}
		return val;
	}
	
	private static int log(int x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static int log(long x, int base) {
		return (int) (Math.log(x) / Math.log(base));
	}
	private static long max(long a, long b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static long abs(long a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int abs(int a) {
		if (a < 0) {
			return -a;
		}
		return a;
	}

	private static int max(int a, int b) {
		if (a >= b) {
			return a;
		}
		return b;
	}

	private static int min(int a, int b) {
		if (a <= b) {
			return a;
		}
		return b;
	}
	
	private static long min(long a, long b) {
		if (a <= b) {
			return a;
		}
		return b;
	}

	private static long gcd(long a, long b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static int gcd(int a, int b) {
		if (b == 0) {
			return a;
		}
		return gcd(b, a % b);
	}

	private static long bigMod(long n, long k, long m) {

		long ans = 1;
		while (k > 0) {
			if ((k & 1) == 1) {
				ans = (ans * n) % m;
			}
			n = (n * n) % m;
			k >>= 1;
		}
		return ans;
	}

	/*
	 * Returns an iterator pointing to the first element in the range [first,
	 * last] which does not compare less than val.
	 * 
	 */
	private static int lowerBoundNew(long[] arr, long num) {
		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid >= 0 && arr[mid] == num) {
					mid--;
				}
				return mid + 1;
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	/*
	 * upper_bound() is a standard library function in C++ defined in the header
	 * . It returns an iterator pointing to the first element in the range
	 * [first, last) that is greater than value, or last if no such element is
	 * found
	 * 
	 */
	private static int upperBoundNew(long[] arr, long num) {

		int start = 0;
		int end = arr.length - 1;
		int index = 0;
		int len = arr.length;
		int mid = 0;
		while (true) {
			if (start > end) {
				break;
			}
			mid = (start + end) / 2;
			if (arr[mid] > num) {
				end = mid - 1;
			} else if (arr[mid] < num) {
				start = mid + 1;
			} else {
				while (mid < len && arr[mid] == num) {
					mid++;
				}
				if (mid == len - 1 && arr[mid] == num) {
					return mid + 1;
				} else {
					return mid;
				}
			}
		}
		if (arr[mid] < num) {
			return mid + 1;
		}
		return mid;
	}

	private static class InputReader {
		public BufferedReader reader;
		public StringTokenizer tokenizer;

		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream));
			tokenizer = null;
		}

		public String next() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());

				}
			} catch (IOException e) {
				return null;
			}
			return tokenizer.nextToken();
		}

		public String nextLine() {
			String line = null;
			try {
				tokenizer = null;
				line = reader.readLine();
			} catch (IOException e) {
				throw new RuntimeException(e);
			}
			return line;
		}

		public int nextInt() {
			return Integer.parseInt(next());
		}

		public double nextDouble() {
			return Double.parseDouble(next());
		}

		public long nextLong() {
			return Long.parseLong(next());
		}

		public boolean hasNext() {
			try {
				while (tokenizer == null || !tokenizer.hasMoreTokens()) {
					tokenizer = new StringTokenizer(reader.readLine());
				}
			} catch (Exception e) {
				return false;
			}
			return true;
		}
	}
}