#include <bits/stdc++.h>
using namespace std;
#define FOR(i,k,n) for(int i = (int)(k); i < (int)(n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) a.begin(), a.end()
#define MS(m,v) memset(m,v,sizeof(m))
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
const int MOD = 1e9 + 7;
template<class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<class T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class T>
istream& operator >> (istream& is, vector<T>& v)
{
	for (auto &i : v) is >> i;
	return is;
}
template<class T>
ostream& operator<<(ostream& os, vector<T>& v)
{
	const string delimiter = "\n";
	REP(i, v.size())
	{
		os << v[i];
		if (i != v.size() - 1) os << delimiter;
	}
	return os;
}
/*--------------------template--------------------*/

int main()
{
	cin.sync_with_stdio(false); cout << fixed << setprecision(10);
	int n;
	cin >> n;
	vector<vi> v(3);
	REP(i, n)
	{
		int a; cin >> a;
		if (a == 2) v[0].push_back(i+1);
		if (a == -2) v[1].push_back(i+1);
		if (a == -1) v[2].push_back(i + 1);
	}
	vi ans;
	for (auto i : v[0]) ans.push_back(i);
	if (v[1].size() % 2 == 0)
	{
		for (auto i : v[1]) ans.push_back(i);
	}
	else
	{
		if (v[2].size() > 0)
		{
			for (auto i : v[1]) ans.push_back(i);
			ans.push_back(v[2][0]);
		}
		else
		{
			for (auto i : v[1]) ans.push_back(i);
			ans.pop_back();
		}
	}
	sort(ALL(ans));
	cout << ans.size() << endl;
	if(ans.size() > 0) cout << ans << endl;
	return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 <bits/stdc++.h>
using namespace std;
#define FOR(i,k,n) for(int i = (int)(k); i < (int)(n); i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(a) a.begin(), a.end()
#define MS(m,v) memset(m,v,sizeof(m))
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int, int> pii;
const int MOD = 1e9 + 7;
template<class T> T &chmin(T &a, const T &b) { return a = min(a, b); }
template<class T> T &chmax(T &a, const T &b) { return a = max(a, b); }
template<class T>
istream& operator >> (istream& is, vector<T>& v)
{
	for (auto &i : v) is >> i;
	return is;
}
template<class T>
ostream& operator<<(ostream& os, vector<T>& v)
{
	const string delimiter = "\n";
	REP(i, v.size())
	{
		os << v[i];
		if (i != v.size() - 1) os << delimiter;
	}
	return os;
}
/*--------------------template--------------------*/

int main()
{
	cin.sync_with_stdio(false); cout << fixed << setprecision(10);
	int n;
	cin >> n;
	vector<vi> v(3);
	REP(i, n)
	{
		int a; cin >> a;
		if (a == 2) v[0].push_back(i+1);
		if (a == -2) v[1].push_back(i+1);
		if (a == -1) v[2].push_back(i + 1);
	}
	vi ans;
	for (auto i : v[0]) ans.push_back(i);
	if (v[1].size() % 2 == 0)
	{
		for (auto i : v[1]) ans.push_back(i);
	}
	else
	{
		if (v[2].size() > 0)
		{
			for (auto i : v[1]) ans.push_back(i);
			ans.push_back(v[2][0]);
		}
		else
		{
			for (auto i : v[1]) ans.push_back(i);
			ans.pop_back();
		}
	}
	sort(ALL(ans));
	cout << ans.size() << endl;
	if(ans.size() > 0) cout << ans << endl;
	return 0;
}

```