    #include <bits/stdc++.h>
     
    using namespace std;
     
    typedef long long ll;
     
    int n1 = 0, n2 = 0, n = 0, s1[200], s2[200];
     
    void work(ll N) {
    	if(N == 1) return;
    	work(N >> 1);
    	s2[++n2] = ++n;
    	if(N & 1) s1[++n1] = ++n;
    }
     
    int main() {
    	ll N; scanf("%lld", &N); work(++N);
    	printf("%d\n", n << 1);
    	for(int i = 1; i <= n; i++) printf("%d ", i);
    	for(int i = n1; i >= 1; i--) printf("%d ", s1[i]);
    	for(int i = 1; i <= n2; i++) printf("%d ", s2[i]);
    	puts("");
    	return 0;
    }
### Prompt

Develop a solution in Cpp to the problem described below:

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 n1 = 0, n2 = 0, n = 0, s1[200], s2[200];
     
    void work(ll N) {
    	if(N == 1) return;
    	work(N >> 1);
    	s2[++n2] = ++n;
    	if(N & 1) s1[++n1] = ++n;
    }
     
    int main() {
    	ll N; scanf("%lld", &N); work(++N);
    	printf("%d\n", n << 1);
    	for(int i = 1; i <= n; i++) printf("%d ", i);
    	for(int i = n1; i >= 1; i--) printf("%d ", s1[i]);
    	for(int i = 1; i <= n2; i++) printf("%d ", s2[i]);
    	puts("");
    	return 0;
    }
```