#include <iostream>
#include <vector>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	vector<int> res;
	int mn = -10;
	int mni = -1;
	int nc = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		if (A[i] != 0 && A[i] != 1) {
			res.push_back(i);
			if (A[i] < 0) {
				nc++;
				if (mn <= A[i]) {
					mn = A[i];
					mni = (int)res.size() - 1;
				}
			}
		}
	}
	if (nc % 2 == 1) {
		res.erase(res.begin() + mni);
	}
	int h = -1;

	for (int i = (int)res.size() - 1; i >= 0; i--) {
		//cerr << i << " " << res[i] << endl;
		if (A[res[i]] == -1) {
			if (h == -1) {
				h = i;
			}
			else {
				res.erase(res.begin() + h);
				res.erase(res.begin() + i);
				h = -1;
			}
		}
	}
	cout << res.size() << endl;
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] + 1 << endl;
	}
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N;
	cin >> N;
	vector<int> A(N);
	vector<int> res;
	int mn = -10;
	int mni = -1;
	int nc = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		if (A[i] != 0 && A[i] != 1) {
			res.push_back(i);
			if (A[i] < 0) {
				nc++;
				if (mn <= A[i]) {
					mn = A[i];
					mni = (int)res.size() - 1;
				}
			}
		}
	}
	if (nc % 2 == 1) {
		res.erase(res.begin() + mni);
	}
	int h = -1;

	for (int i = (int)res.size() - 1; i >= 0; i--) {
		//cerr << i << " " << res[i] << endl;
		if (A[res[i]] == -1) {
			if (h == -1) {
				h = i;
			}
			else {
				res.erase(res.begin() + h);
				res.erase(res.begin() + i);
				h = -1;
			}
		}
	}
	cout << res.size() << endl;
	for (int i = 0; i < res.size(); i++) {
		cout << res[i] + 1 << endl;
	}
}

```