#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;

int main(void) {
  int i,j;
  int n;
  cin>>n;
  vi a(n);
  rep(i,n)cin>>a[i];
  int x=0,y=0,z=0;
  rep(i,n){
    if(a[i]==2)x++;
    if(a[i]==-2)y++;
    if(a[i]==-1)z++;
  }

  //  cout<<x<<" "<<y<<" "<<z<<endl;

  if(y%2==1){
    if(z){
      cout<<x+y+1<<endl;
      rep(i,n){
	if(a[i]==2 || a[i]==-2)cout<<i+1<<endl;
	if(a[i]==-1 && z){
	  cout<<i+1<<endl;
	  z=0;
	}
      }
    }else{
	cout<<x+y-1<<endl;
	rep(i,n){
	  if(a[i]==2)cout<<i+1<<endl;
	  if(a[i]==-2 && y-1>0){
	    cout<<i+1<<endl;
	    y--;
	  }
	}

    }

  }else{
    cout<<x+y<<endl;
    rep(i,n)if(a[i]==2 || a[i]==-2)cout<<i+1<<endl;

  }

}


### 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<iostream>
#include<algorithm>
#include<cstdio>
#include<cmath>
#include<cctype>
#include<math.h>
#include<string>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<utility>
#include<set>
#include<map>
#include<stdlib.h>
#include<iomanip>

using namespace std;

#define ll long long
#define ld long double
#define EPS 0.0000000001
#define INF 1e9
#define MOD 1000000007
#define rep(i,n) for(i=0;i<(n);i++)
#define loop(i,a,n) for(i=a;i<(n);i++)
#define all(in) in.begin(),in.end()
#define shosu(x) fixed<<setprecision(x)

typedef vector<int> vi;
typedef vector<string> vs;
typedef pair<int,int> pii;

int main(void) {
  int i,j;
  int n;
  cin>>n;
  vi a(n);
  rep(i,n)cin>>a[i];
  int x=0,y=0,z=0;
  rep(i,n){
    if(a[i]==2)x++;
    if(a[i]==-2)y++;
    if(a[i]==-1)z++;
  }

  //  cout<<x<<" "<<y<<" "<<z<<endl;

  if(y%2==1){
    if(z){
      cout<<x+y+1<<endl;
      rep(i,n){
	if(a[i]==2 || a[i]==-2)cout<<i+1<<endl;
	if(a[i]==-1 && z){
	  cout<<i+1<<endl;
	  z=0;
	}
      }
    }else{
	cout<<x+y-1<<endl;
	rep(i,n){
	  if(a[i]==2)cout<<i+1<<endl;
	  if(a[i]==-2 && y-1>0){
	    cout<<i+1<<endl;
	    y--;
	  }
	}

    }

  }else{
    cout<<x+y<<endl;
    rep(i,n)if(a[i]==2 || a[i]==-2)cout<<i+1<<endl;

  }

}


```