Problem: 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.

Examples

Input

7


Output

4
1 1 1 1


Input

299


Output

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 CPP):

Solution:
#include<cstdio>
#include<deque>
#define LL long long
using namespace std;
typedef deque<int>::iterator diter;
LL n;
int cnt;
deque<int> x,y;
void solve(LL n)
{
	if(n==1) return;
	solve(n/2);
	x.push_front(++cnt); y.push_front(cnt);
	if(n&1) { x.push_front(++cnt); y.push_back(cnt); }
}
int main()
{
	scanf("%lld",&n); n++;
	solve(n);
	printf("%d\n",cnt*2);
	for(diter it=x.begin();it!=x.end();it++) printf("%d ",*it);
	for(diter it=y.begin();it!=y.end();it++) printf("%d ",*it);
}