#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, a[100], P, Q, R; vector<int>S[5];
long long maxn = -(1LL << 60); vector<int>maxid;

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		S[a[i] + 2].push_back(i);
	}
	for (int i = 0; i <= S[0].size(); i++) {
		for (int j = 0; j <= S[1].size(); j++) {
			for (int k = 0; k <= S[2].size(); k++) {
				for (int l = 0; l <= S[3].size(); l++) {
					for (int m = 0; m <= S[4].size(); m++) {
						long long ret = 1; vector<int>A;
						for (int p = 0; p < i; p++) { ret *= -2; A.push_back(S[0][p]); }
						for (int p = 0; p < j; p++) { ret *= -1; A.push_back(S[1][p]); }
						for (int p = 0; p < k; p++) { ret *= 0; A.push_back(S[2][p]); }
						for (int p = 0; p < l; p++) { ret *= 1; A.push_back(S[3][p]); }
						for (int p = 0; p < m; p++) { ret *= 2; A.push_back(S[4][p]); }
						sort(A.begin(), A.end());
						if (ret > maxn) {
							maxn = ret;
							maxid = A;
						}
						else if (ret == maxn) {
							if (A.size() < maxid.size()) maxid = A;
							else if (A.size() == maxid.size() && A < maxid) maxid = A;
						}
					}
				}
			}
		}
	}
	cout << maxid.size() << endl;
	for (int i = 0; i < maxid.size(); i++) cout << maxid[i] << endl;
	return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

B： 階層的計算機 (Hierarchical Calculator)

Problem

Ebi-chan has N formulae: y = a_i x for i =1, ..., N (inclusive). Now she considers a subsequence of indices with length k: s_1, s_2, ..., s_k. At first, let x_0 be 1 and evaluate s_1-th formulae with x = x_0. Next, let x_1 be the output of s_1 and evaluate s_2-th formulae with x = x_1, and so on.

She wants to maximize the final output of the procedure, x_{s_k}. If there are many candidates, she wants the """shortest one""". If there are still many candidates, she wants the """lexicographically smallest one""".

Sequence s is lexicographically smaller than sequence t, if and only if either of the following conditions hold:

* there exists m < |s| such that s_i = t_i for i in 1 to m (inclusive) and s_{m+1} < t_{m+1}, or
* s_i = t_i for i in 1 to |s| (inclusive) and |s| < |t|,



where |s| is the length of the s.

Input


N
a_1 a_2 $\cdots$ a_N


Constraints

* 1 \leq N \leq 60
* -2 \leq a_i \leq 2 for i=1, ...,N (inclusive)
* Every input is given as the integer.



Output

Output k+1 lines. First line, the length of the sequence, k. Following k lines, the index of the i-th element of the subsequence, s_i (one element per line).

Sample Input 1


4
2 0 -2 1


Sample Output for Input 1


1
1


She evaluates the first one and gets the maximum value 2.

Sample Input 2


3
2 -2 -2


Sample Output for Input 2


3
1
2
3


She evaluates all of them and gets the maximum value 8.

Sample Input 3


2
-1 0


Sample Output for Input 3


0

She evaluates none of them and gets the maximum value 0. Empty sequence is the shorter and lexicographically smaller than any other sequences.

Sample Input 4


5
-1 2 1 -2 -1


Sample Output for Input 4


3
1
2
4


She evaluates $\langle$ 1, 2, 4 $\rangle$ ones and gets the maximum value 4. Note that $\langle$ 2, 4, 5 $\rangle$ is not lexicographically smallest one.





Example

Input

4
2 0 -2 1


Output

1
1

### Response

```cpp
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int n, a[100], P, Q, R; vector<int>S[5];
long long maxn = -(1LL << 60); vector<int>maxid;

int main() {
	cin >> n;
	for (int i = 1; i <= n; i++) {
		cin >> a[i];
		S[a[i] + 2].push_back(i);
	}
	for (int i = 0; i <= S[0].size(); i++) {
		for (int j = 0; j <= S[1].size(); j++) {
			for (int k = 0; k <= S[2].size(); k++) {
				for (int l = 0; l <= S[3].size(); l++) {
					for (int m = 0; m <= S[4].size(); m++) {
						long long ret = 1; vector<int>A;
						for (int p = 0; p < i; p++) { ret *= -2; A.push_back(S[0][p]); }
						for (int p = 0; p < j; p++) { ret *= -1; A.push_back(S[1][p]); }
						for (int p = 0; p < k; p++) { ret *= 0; A.push_back(S[2][p]); }
						for (int p = 0; p < l; p++) { ret *= 1; A.push_back(S[3][p]); }
						for (int p = 0; p < m; p++) { ret *= 2; A.push_back(S[4][p]); }
						sort(A.begin(), A.end());
						if (ret > maxn) {
							maxn = ret;
							maxid = A;
						}
						else if (ret == maxn) {
							if (A.size() < maxid.size()) maxid = A;
							else if (A.size() == maxid.size() && A < maxid) maxid = A;
						}
					}
				}
			}
		}
	}
	cout << maxid.size() << endl;
	for (int i = 0; i < maxid.size(); i++) cout << maxid[i] << endl;
	return 0;
}

```