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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#if 0
date +%Y.%m.%d

醉了，一开始有个 sb 想法，结果最坏 s 的长度可以被卡到 700+ ，就不说了。
构造，s 前半部分是 1 到 p 的排列，后半部分是 1 到 p ，
那么
#endif
#include <cstdio>
#define debug(...) fprintf(stderr, __VA_ARGS__)

typedef long long ll;
typedef unsigned long long ull;

struct {
	inline operator ll () { ll x; return scanf("%lld", &x), x; }
	template<class T> inline void operator () (T &x) { x = *this; }
	template<class T, class ...A> inline void operator () (T &x, A &...a)
	{ x = *this; this -> operator ()(a...); }
} read;

int a[200], al = 100, ar = 100;
int p = 1;
void work(ll n) {
	if(!n) return;
	if(n & 1) {
		a[-- al] = ++ p;
		work(n ^ 1);
	} else {
		a[++ ar] = ++ p;
		work(n >> 1);
	}
}

int main() {
	a[100] = 1;
	work(read);
	printf("%d\n", p << 1);
	for(int i = al; i <= ar; i ++)
		printf("%d ", a[i]);
	for(int i = 1; i <= p; i ++)
		printf("%d ", i);
	puts("");
}