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
#include<iostream>
#include<string>
#include<iomanip>
#include<cmath>
#include<vector>
#include<algorithm>

using namespace std;

#define int long long
#define rep(i,n) for(int i = 0; i < (n); i++)
#define INF ((long long)1e18)
#define MOD ((int)1e9+7)
#define endl "\n"

#define yn(f) ((f)?"Yes":"No")
#define YN(f) ((f)?"YES":"NO")

#define MAX 100

signed main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cout<<fixed<<setprecision(10);
	
	int N, a[MAX], used[MAX] = {};
	int num = 1, con = 0;
	
	cin>>N;
	
	for(int i = 1; i <= N; i++){
		cin>>a[i];
		if(a[i] == 2 || a[i] == -2) used[i] = true, con++;
		if(a[i] == -2) num *= -1;
	}
	
	if(num == -1){
		bool flag = false;
		
		for(int i = 1; i <= N; i++){
			if(a[i] == -1){
				used[i] = true;
				flag = true;
				con++;
				break;
			}
		}
		
		if(flag == false){
			
			for(int i = N; i; i--){
				if(a[i] == -2){
					used[i] = false;
					con--;
					break;
				}
			}
			
		}
	}
	
	cout<<con<<endl;
	
	for(int i = 1; i <= N; i++){
		if(used[i])cout<<i<<endl;
	}
	
	return 0;
}
