#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
using ll = long long;
ll dp[61][2];
vector<int> ag[61][2];
bool better(vector<int> a,vector<int> b){
	if(a.size()!=b.size()) return a.size()<b.size();
	return a<b;
}
int main(){
	int N;
	cin>>N;
	dp[0][0] = dp[0][1] = 1;
	rep(i,N){
		int a;
		cin>>a;
		rep(j,2) dp[i+1][j] = dp[i][j], ag[i+1][j] = ag[i][j];
		rep(j,2){
			rep(t,2){
				ll nxt = dp[i][j]*a;
				ll now = dp[i+1][t];
				vector<int> nag = ag[i][j];
				nag.pb(i);
				if(t==0){
					if(nxt<now || (nxt==now && better(nag,ag[i+1][t]))){
						dp[i+1][t] = nxt;
						ag[i+1][t] = nag;
					}
				}else{
					if(nxt>now || (nxt==now && better(nag,ag[i+1][t]))){
						dp[i+1][t] = nxt;
						ag[i+1][t] = nag;
					}
				}
			}
		}
	}
	vector<int> ans = ag[N][1];
	cout<<ans.size()<<endl;
	for(int v:ans) cout<<v+1<<endl;
}


### Prompt

Develop a solution in cpp 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 <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define all(c) c.begin(),c.end()
#define pb push_back
#define fs first
#define sc second
#define show(x) cout << #x << " = " << x << endl
#define chmin(x,y) x=min(x,y)
#define chmax(x,y) x=max(x,y)
using namespace std;
template<class S,class T> ostream& operator<<(ostream& o,const pair<S,T> &p){return o<<"("<<p.fs<<","<<p.sc<<")";}
template<class T> ostream& operator<<(ostream& o,const vector<T> &vc){o<<"sz = "<<vc.size()<<endl<<"[";for(const T& v:vc) o<<v<<",";o<<"]";return o;}
using ll = long long;
ll dp[61][2];
vector<int> ag[61][2];
bool better(vector<int> a,vector<int> b){
	if(a.size()!=b.size()) return a.size()<b.size();
	return a<b;
}
int main(){
	int N;
	cin>>N;
	dp[0][0] = dp[0][1] = 1;
	rep(i,N){
		int a;
		cin>>a;
		rep(j,2) dp[i+1][j] = dp[i][j], ag[i+1][j] = ag[i][j];
		rep(j,2){
			rep(t,2){
				ll nxt = dp[i][j]*a;
				ll now = dp[i+1][t];
				vector<int> nag = ag[i][j];
				nag.pb(i);
				if(t==0){
					if(nxt<now || (nxt==now && better(nag,ag[i+1][t]))){
						dp[i+1][t] = nxt;
						ag[i+1][t] = nag;
					}
				}else{
					if(nxt>now || (nxt==now && better(nag,ag[i+1][t]))){
						dp[i+1][t] = nxt;
						ag[i+1][t] = nag;
					}
				}
			}
		}
	}
	vector<int> ans = ag[N][1];
	cout<<ans.size()<<endl;
	for(int v:ans) cout<<v+1<<endl;
}


```