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: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <math.h>
#include <limits.h>
#define swap(type,a,b) do{type t=a;a=b;b=t;}while(0);
#define MAX(a,b) (((a)>(b))?(a):(b))
#define MIN(a,b) (((a)<(b))?(a):(b))
#define ll long long
#define INF 100000000
#define MOD 1000000007
#define SIZE 100005
#define FOR(i,a,n) for(i=(a);i<(n);i++)
int comp(const void* a,const void* b){
	return *(int*)a-*(int*)b;
}
int res[200];
int add[200];
int main(void)
{
	ll n;
	int m,size,s=0,i;
	scanf("%lld",&n);
	for(m=50;n<(1LL<<m)-1;m--);
	//printf("%d\n",1<<m);
	FOR(i,1,m+1) res[i-1]=i;
	n-=(1LL<<m)-1;
	size=m;
	for(i=m-1;i>=0;i--){
		if(n<(1LL<<i)) continue;
		m++;
		add[i]=m;
		n-=(1LL<<i);
	}
	printf("%d\n",m*2);
	FOR(i,0,size){
		if(add[i]) printf("%d ",add[i]);
		printf("%d ",res[i]);
	}
	FOR(i,0,m) printf("%d%c",i+1,i==m-1?'\n':' ');
	return 0;
}