#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

int main(){
    int n;
    cin >>n;
    vector<int> a(n);
    rep(i,n) cin >>a[i];

    int p=0,m=0;
    int o=0;
    rep(i,n){
        if(a[i]==2) ++p;
        else if(a[i]==-2) ++m;
        else if(a[i]==-1) ++o;
    }

    if(m%2==1){
        if(o==0) --m;
        else o=1;
    }
    else{
        o = 0;
    }

    vector<int> ans;
    rep(i,n){
        if(a[i]==2){
            if(p){
                ans.pb(i+1);
                --p;
            }
        }
        else if(a[i]==-2){
            if(m){
                ans.pb(i+1);
                --m;
            }
        }
        else if(a[i]==-1){
            if(o){
                ans.pb(i+1);
                --o;
            }
        }
    }

    cout << ans.size() << endl;
    for(int i:ans) cout << i << endl;
    return 0;
}


### 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>
using namespace std;
using ll = long long;
#define rep(i,n) for(int (i)=0;(i)<(int)(n);++(i))
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define fi first
#define se second
#define dbg(x) cout<<#x" = "<<((x))<<endl
template<class T,class U> ostream& operator<<(ostream& o, const pair<T,U> &p){o<<"("<<p.fi<<","<<p.se<<")";return o;}
template<class T> ostream& operator<<(ostream& o, const vector<T> &v){o<<"[";for(T t:v){o<<t<<",";}o<<"]";return o;}

int main(){
    int n;
    cin >>n;
    vector<int> a(n);
    rep(i,n) cin >>a[i];

    int p=0,m=0;
    int o=0;
    rep(i,n){
        if(a[i]==2) ++p;
        else if(a[i]==-2) ++m;
        else if(a[i]==-1) ++o;
    }

    if(m%2==1){
        if(o==0) --m;
        else o=1;
    }
    else{
        o = 0;
    }

    vector<int> ans;
    rep(i,n){
        if(a[i]==2){
            if(p){
                ans.pb(i+1);
                --p;
            }
        }
        else if(a[i]==-2){
            if(m){
                ans.pb(i+1);
                --m;
            }
        }
        else if(a[i]==-1){
            if(o){
                ans.pb(i+1);
                --o;
            }
        }
    }

    cout << ans.size() << endl;
    for(int i:ans) cout << i << endl;
    return 0;
}


```