#include <cstdio>
typedef long long ll;
const int N = 205;
ll n;
int a[55], len, cnt, length, lx[N], rx[N], nl, nr;
void solve(ll x) {
	while (x) a[++len] = x & 1, x >>= 1;
	nl = nr = 0;
	for (int i = len - 1; i; --i) {
		lx[++nl] = ++cnt;
		if (a[i]) rx[++nr] = ++cnt;
	}
		
	printf("%d\n", cnt << 1);
	for (int i = cnt; i; --i) printf("%d ", i);
	for (int i = nl; i; --i) printf("%d ", lx[i]);
	for (int i = 1; i <= nr; ++i) printf("%d ", rx[i]);
}
int main() {
	scanf("%lld", &n);
	solve(n + 1);
}
### 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>
typedef long long ll;
const int N = 205;
ll n;
int a[55], len, cnt, length, lx[N], rx[N], nl, nr;
void solve(ll x) {
	while (x) a[++len] = x & 1, x >>= 1;
	nl = nr = 0;
	for (int i = len - 1; i; --i) {
		lx[++nl] = ++cnt;
		if (a[i]) rx[++nr] = ++cnt;
	}
		
	printf("%d\n", cnt << 1);
	for (int i = cnt; i; --i) printf("%d ", i);
	for (int i = nl; i; --i) printf("%d ", lx[i]);
	for (int i = 1; i <= nr; ++i) printf("%d ", rx[i]);
}
int main() {
	scanf("%lld", &n);
	solve(n + 1);
}
```