#include<bits/stdc++.h>
#define F first
#define S second
#define int long long
using namespace std;

typedef pair<int,int> P;
double eps=1e-8;
int mod=1e9+7;
int inf=1e9;

priority_queue<P,vector<P>,greater<P> > pq;

template <typename T>
int Count(const vector<T> &v, T l,T r){
  //counting [l,r]
  auto L = lower_bound(v.begin(), v.end() , l );
  auto R = upper_bound(v.begin(), v.end(), r );
  // [l,r) change upper -> lower
  return R-L;
}

signed main(){
  vector<P> v[2];
  int n,a[60],c=0;
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];
    if(a[i]>0)v[0].push_back(P(-a[i],i));
    if(a[i]<0)v[1].push_back(P(a[i],i));
    if(a[i]==-2)c++;
  }

  sort(v[0].begin(),v[0].end());
  sort(v[1].begin(),v[1].end());
  vector<int> ans;
  if(c%2){
    if(v[1].size()==c) c--;
    else c++;
  }
  
  for(int i=0;i<v[0].size();i++)if(v[0][i].first==-2)ans.push_back(v[0][i].second);
  for(int i=0;i<v[1].size();i++)if(c)ans.push_back(v[1][i].second),c--;
  sort(ans.begin(),ans.end());
  cout<<ans.size()<<endl;
  for(int i=0;i<ans.size();i++)cout<<ans[i]+1<<endl;
  return 0;
}


### Prompt

Please formulate 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<bits/stdc++.h>
#define F first
#define S second
#define int long long
using namespace std;

typedef pair<int,int> P;
double eps=1e-8;
int mod=1e9+7;
int inf=1e9;

priority_queue<P,vector<P>,greater<P> > pq;

template <typename T>
int Count(const vector<T> &v, T l,T r){
  //counting [l,r]
  auto L = lower_bound(v.begin(), v.end() , l );
  auto R = upper_bound(v.begin(), v.end(), r );
  // [l,r) change upper -> lower
  return R-L;
}

signed main(){
  vector<P> v[2];
  int n,a[60],c=0;
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];
    if(a[i]>0)v[0].push_back(P(-a[i],i));
    if(a[i]<0)v[1].push_back(P(a[i],i));
    if(a[i]==-2)c++;
  }

  sort(v[0].begin(),v[0].end());
  sort(v[1].begin(),v[1].end());
  vector<int> ans;
  if(c%2){
    if(v[1].size()==c) c--;
    else c++;
  }
  
  for(int i=0;i<v[0].size();i++)if(v[0][i].first==-2)ans.push_back(v[0][i].second);
  for(int i=0;i<v[1].size();i++)if(c)ans.push_back(v[1][i].second),c--;
  sort(ans.begin(),ans.end());
  cout<<ans.size()<<endl;
  for(int i=0;i<ans.size();i++)cout<<ans[i]+1<<endl;
  return 0;
}


```