#include <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <set>
#include <map>
#include <time.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

#define NUM 60

int main(){

	int N;
	scanf("%d",&N);

	int table[N];

	int num_2 = 0,num_neg2 = 0,num_neg1 = 0;

	for(int i = 0; i < N; i++){
		scanf("%d",&table[i]);

		switch(table[i]){
		case 2:
			num_2++;
			break;
		case -2:
			num_neg2++;
			break;
		case -1:
			num_neg1++;
			break;
		default:
			//Do nothing
			break;
		}
	}

	if(num_neg2%2 == 0){
		num_neg1 = 0;
	}else{
		if(num_neg1 == 0){
			num_neg2 -= 1;
		}else{
			num_neg1 = 1;
		}
	}

	printf("%d\n",num_2+num_neg2+num_neg1);

	for(int i = 0; i < N; i++){
		switch(table[i]){
		case 2:
			printf("%d\n",i+1);
			break;
		case -2:
			if(num_neg2 > 0){
				printf("%d\n",i+1);
				num_neg2--;
			}
			break;
		case -1:
			if(num_neg1 > 0){
				printf("%d\n",i+1);
				num_neg1--;
			}
			break;
		default:
			//Do nothing
			break;
		}
	}

	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 <stdio.h>
#include <cmath>
#include <algorithm>
#include <cfloat>
#include <stack>
#include <queue>
#include <vector>
#include <string>
#include <iostream>
#include <set>
#include <map>
#include <time.h>
typedef long long int ll;
typedef unsigned long long int ull;
#define BIG_NUM 2000000000
#define MOD 1000000007
#define EPS 0.000000001
using namespace std;

#define NUM 60

int main(){

	int N;
	scanf("%d",&N);

	int table[N];

	int num_2 = 0,num_neg2 = 0,num_neg1 = 0;

	for(int i = 0; i < N; i++){
		scanf("%d",&table[i]);

		switch(table[i]){
		case 2:
			num_2++;
			break;
		case -2:
			num_neg2++;
			break;
		case -1:
			num_neg1++;
			break;
		default:
			//Do nothing
			break;
		}
	}

	if(num_neg2%2 == 0){
		num_neg1 = 0;
	}else{
		if(num_neg1 == 0){
			num_neg2 -= 1;
		}else{
			num_neg1 = 1;
		}
	}

	printf("%d\n",num_2+num_neg2+num_neg1);

	for(int i = 0; i < N; i++){
		switch(table[i]){
		case 2:
			printf("%d\n",i+1);
			break;
		case -2:
			if(num_neg2 > 0){
				printf("%d\n",i+1);
				num_neg2--;
			}
			break;
		case -1:
			if(num_neg1 > 0){
				printf("%d\n",i+1);
				num_neg1--;
			}
			break;
		default:
			//Do nothing
			break;
		}
	}

	return 0;
}


```