#include<algorithm>
#include<iostream>
#include<vector>
using namespace std;
typedef long long lint;
typedef vector<int>vi;
typedef pair<int,int>pii;
typedef pair<lint,lint>pll;
#define rep(i,n)for(int i=0;i<(int)(n);++i)

const int DEBUG=0;


const int N=62;
const lint inf=1LL<<61;
pll dpa[N],dpb[N];

pll mul(pll a,lint b){
  return pll(a.first*b,a.second+1);
}

pll rev(pll a){
  return pll(a.first,-a.second);
}

const int PUL=0;


int main(){
  int n;
  cin>>n;
  vector<lint>a(n);
  if(PUL){
    rep(i,n)a[i]=-2;
  }else{
    rep(i,n)cin>>a[i];
  }
  rep(i,N){
    dpa[i]=pll(-inf,1);
    dpb[i]=pll(inf,-1);
  }
  dpa[n]=dpb[n]=pll(1,0);
  for(int i=n-1;i>=0;--i){
    for(int j=i+1;j<=n;++j){
      dpa[i]=max(dpa[i],max(rev(mul(rev(dpa[j]),a[i])),rev(mul(dpb[j],a[i]))));
      dpb[i]=min(dpb[i],min(mul(rev(dpa[j]),a[i]),mul(dpb[j],a[i])));
    }
    if(DEBUG){
      cerr<<"dpa["<<i<<"]="<<dpa[i].first<<" "<<dpa[i].second<<endl;
      cerr<<"dpb["<<i<<"]="<<dpb[i].first<<" "<<dpb[i].second<<endl;
    }
  }
  pll ma=pll(-inf,1);
  rep(i,n+1)ma=max(ma,dpa[i]);
  pll idx(inf,inf);
  rep(i,n+1){
    if(ma.first==dpa[i].first){
      idx=min(idx,pll(-dpa[i].second,i));
    }
  }
  int cur=idx.second;
  int rem=-ma.second;
  pll val=rev(ma);
  vi ans;
  while(rem>0){
    if(DEBUG){
      cerr<<"cur="<<cur<<endl;
      cerr<<"rem="<<rem<<endl;
      cerr<<"val="<<val.first<<" "<<val.second<<endl;
    }
    int nxt=-1;
    for(int j=cur+1;j<=n;++j){
      if(mul(dpb[j],a[cur])==val){
	nxt=j;
	val=dpb[j];
	break;
      }
      if(mul(rev(dpa[j]),a[cur])==val){
	nxt=j;
	val=rev(dpa[j]);
	break;
      }
    }
    ans.push_back(cur);
    cur=nxt;
    rem--;
  }
  cout<<ans.size()<<endl;
  rep(i,ans.size()){
    cout<<ans[i]+1<<endl;
  }
}


### Prompt

Create a solution in CPP for 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<algorithm>
#include<iostream>
#include<vector>
using namespace std;
typedef long long lint;
typedef vector<int>vi;
typedef pair<int,int>pii;
typedef pair<lint,lint>pll;
#define rep(i,n)for(int i=0;i<(int)(n);++i)

const int DEBUG=0;


const int N=62;
const lint inf=1LL<<61;
pll dpa[N],dpb[N];

pll mul(pll a,lint b){
  return pll(a.first*b,a.second+1);
}

pll rev(pll a){
  return pll(a.first,-a.second);
}

const int PUL=0;


int main(){
  int n;
  cin>>n;
  vector<lint>a(n);
  if(PUL){
    rep(i,n)a[i]=-2;
  }else{
    rep(i,n)cin>>a[i];
  }
  rep(i,N){
    dpa[i]=pll(-inf,1);
    dpb[i]=pll(inf,-1);
  }
  dpa[n]=dpb[n]=pll(1,0);
  for(int i=n-1;i>=0;--i){
    for(int j=i+1;j<=n;++j){
      dpa[i]=max(dpa[i],max(rev(mul(rev(dpa[j]),a[i])),rev(mul(dpb[j],a[i]))));
      dpb[i]=min(dpb[i],min(mul(rev(dpa[j]),a[i]),mul(dpb[j],a[i])));
    }
    if(DEBUG){
      cerr<<"dpa["<<i<<"]="<<dpa[i].first<<" "<<dpa[i].second<<endl;
      cerr<<"dpb["<<i<<"]="<<dpb[i].first<<" "<<dpb[i].second<<endl;
    }
  }
  pll ma=pll(-inf,1);
  rep(i,n+1)ma=max(ma,dpa[i]);
  pll idx(inf,inf);
  rep(i,n+1){
    if(ma.first==dpa[i].first){
      idx=min(idx,pll(-dpa[i].second,i));
    }
  }
  int cur=idx.second;
  int rem=-ma.second;
  pll val=rev(ma);
  vi ans;
  while(rem>0){
    if(DEBUG){
      cerr<<"cur="<<cur<<endl;
      cerr<<"rem="<<rem<<endl;
      cerr<<"val="<<val.first<<" "<<val.second<<endl;
    }
    int nxt=-1;
    for(int j=cur+1;j<=n;++j){
      if(mul(dpb[j],a[cur])==val){
	nxt=j;
	val=dpb[j];
	break;
      }
      if(mul(rev(dpa[j]),a[cur])==val){
	nxt=j;
	val=rev(dpa[j]);
	break;
      }
    }
    ans.push_back(cur);
    cur=nxt;
    rem--;
  }
  cout<<ans.size()<<endl;
  rep(i,ans.size()){
    cout<<ans[i]+1<<endl;
  }
}


```