#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int q[222];
int cnt=0,l,r;
void work(ll x){
    if(x==0)return ;
    if(x%2==0){
        //int idx=l--;
        work(x-1);
        q[l--]=++cnt;

    }
    else {
        work(x/2);
        q[r++]=++cnt;
    }
}
int main()
{
    ll n;
    scanf("%lld",&n);
    l=100,r=101;
    cnt=0;
    work(n);
    printf("%d\n",cnt*2);
    for(int i=l+1;i<r;i++){
        printf("%d ",q[i]);
    }
    for(int i=1;i<=cnt;i++){
        printf("%d",i);
        if(i!=cnt)printf(" ");
        else printf("\n");
    }
    return 0;
}

### Prompt

Your task is to create a CPP solution to the following 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

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int q[222];
int cnt=0,l,r;
void work(ll x){
    if(x==0)return ;
    if(x%2==0){
        //int idx=l--;
        work(x-1);
        q[l--]=++cnt;

    }
    else {
        work(x/2);
        q[r++]=++cnt;
    }
}
int main()
{
    ll n;
    scanf("%lld",&n);
    l=100,r=101;
    cnt=0;
    work(n);
    printf("%d\n",cnt*2);
    for(int i=l+1;i<r;i++){
        printf("%d ",q[i]);
    }
    for(int i=1;i<=cnt;i++){
        printf("%d",i);
        if(i!=cnt)printf(" ");
        else printf("\n");
    }
    return 0;
}

```