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 C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main(){
	long long int N;
	scanf("%lld", &N);
	int k, digit[64];
	for(k = 0; N != 0; k++){
		digit[k] = N & 0x1;
		N >>= 1;
	}
	int i, left[2 * k - 1];
	for(i = 1; i <= k; i++){
		left[2 * i] = i;
	}
	int j = k;
	for(i = 0; i < k - 1; i++){
		if(digit[k - i - 2] == 1){
			left[2 * k - 3 - 2 * i] = j;
			j++;
		}
		else{
			left[2 * k - 3 - 2 * i] = 0;
		}
	}
	left[0] = j;
	for(i = 0; i < 2 * k - 1; i++){
		if(left[i] != 0){
			printf("%d ", left[i]);
		}
	}
	for(i = 1; i <= j; i++){
		printf("%d ", i);
	}
	return 0;
}