#include<bits/stdc++.h>
using namespace std;
using Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  vector<Int> a(n);
  for(Int i=0;i<n;i++) cin>>a[i];

  Int p1=0,p2=0;
  Int m1=0,m2=0;
  for(Int i=0;i<n;i++){
    if(a[i]==+1) p1++;
    if(a[i]==+2) p2++;
    if(a[i]==-1) m1++;
    if(a[i]==-2) m2++;    
  }
  
  if(m1&&(m2&1)) m1=1;
  else if(m2&1) m1=0,m2--;
  else m1=0;

  if(p2+m2==0){
    cout<<0<<endl;
    return 0;
  }
  vector<Int> ans;
  for(Int i=0;i<n;i++){
    if(a[i]==+1) continue;
    if(a[i]==+2) ans.emplace_back(i);
    if(a[i]==-1) if(m1) ans.emplace_back(i),m1--;
    if(a[i]==-2) if(m2) ans.emplace_back(i),m2--;
  }
  cout<<ans.size()<<endl;
  for(Int x:ans) cout<<x+1<<endl;
  return 0;
}


### Prompt

Please provide a cpp coded solution 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 Int = long long;
template<typename T1,typename T2> inline void chmin(T1 &a,T2 b){if(a>b) a=b;}
template<typename T1,typename T2> inline void chmax(T1 &a,T2 b){if(a<b) a=b;}

//INSERT ABOVE HERE
signed main(){
  Int n;
  cin>>n;
  vector<Int> a(n);
  for(Int i=0;i<n;i++) cin>>a[i];

  Int p1=0,p2=0;
  Int m1=0,m2=0;
  for(Int i=0;i<n;i++){
    if(a[i]==+1) p1++;
    if(a[i]==+2) p2++;
    if(a[i]==-1) m1++;
    if(a[i]==-2) m2++;    
  }
  
  if(m1&&(m2&1)) m1=1;
  else if(m2&1) m1=0,m2--;
  else m1=0;

  if(p2+m2==0){
    cout<<0<<endl;
    return 0;
  }
  vector<Int> ans;
  for(Int i=0;i<n;i++){
    if(a[i]==+1) continue;
    if(a[i]==+2) ans.emplace_back(i);
    if(a[i]==-1) if(m1) ans.emplace_back(i),m1--;
    if(a[i]==-2) if(m2) ans.emplace_back(i),m2--;
  }
  cout<<ans.size()<<endl;
  for(Int x:ans) cout<<x+1<<endl;
  return 0;
}


```