#include <cstdio>
int arr[205], ans[205];
int main()
{
	// freopen("AGC012-C.in", "r", stdin);
	long long n;
	scanf("%lld", &n);
	n++;
	int cnt = 0, len = 0;
	while (n != 1)
	{
		cnt += n & 1;
		arr[len++] = n & 1;
		n >>= 1;
	}
	int l = cnt - 1, r = cnt, cur = 0;
	for (int i = len - 1; i >= 0; i--)
	{
		ans[r++] = cur++;
		if (arr[i])
			ans[l--] = cur++;
	}
	printf("%d\n", r << 1);
	for (int i = 0; i < r << 1; i++)
		printf("%d%c", i < r ? ans[i] + 1 : i - r + 1, " \n"[i + 1 == r << 1]);
	return 0;
}

### Prompt

Generate 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 <cstdio>
int arr[205], ans[205];
int main()
{
	// freopen("AGC012-C.in", "r", stdin);
	long long n;
	scanf("%lld", &n);
	n++;
	int cnt = 0, len = 0;
	while (n != 1)
	{
		cnt += n & 1;
		arr[len++] = n & 1;
		n >>= 1;
	}
	int l = cnt - 1, r = cnt, cur = 0;
	for (int i = len - 1; i >= 0; i--)
	{
		ans[r++] = cur++;
		if (arr[i])
			ans[l--] = cur++;
	}
	printf("%d\n", r << 1);
	for (int i = 0; i < r << 1; i++)
		printf("%d%c", i < r ? ans[i] + 1 : i - r + 1, " \n"[i + 1 == r << 1]);
	return 0;
}

```