#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
bool pass(int v)
{
    if(v == 0 || v == 1) return true;
    return false;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    vector<int> cnt(10, 0), A;
    for(int i = 0; i < n; i++)
    {
        int x; cin >> x;
        A.push_back(x);
        cnt[x + 2]++;
    }
    vector<int> ans;
    if(cnt[0] % 2 == 0)
    {
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] ==  -1) continue;
            ans.push_back(i);
        }
    }
    else if(cnt[1] > 0)
    {
        bool flag = false;
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] == -1 && flag) continue;
            if(A[i] == -1) flag = true;
            ans.push_back(i);
        }
    }
    else
    {
        int num = cnt[0] - 1;
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] == -1) continue;
            if(num > 0 && A[i] == -2)
            {
                num--;
                ans.push_back(i);
            }
            else if(A[i] != -2)
            {
                ans.push_back(i);
            }
        }
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++) cout << ans[i] + 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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
bool pass(int v)
{
    if(v == 0 || v == 1) return true;
    return false;
}
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    vector<int> cnt(10, 0), A;
    for(int i = 0; i < n; i++)
    {
        int x; cin >> x;
        A.push_back(x);
        cnt[x + 2]++;
    }
    vector<int> ans;
    if(cnt[0] % 2 == 0)
    {
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] ==  -1) continue;
            ans.push_back(i);
        }
    }
    else if(cnt[1] > 0)
    {
        bool flag = false;
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] == -1 && flag) continue;
            if(A[i] == -1) flag = true;
            ans.push_back(i);
        }
    }
    else
    {
        int num = cnt[0] - 1;
        for(int i = 0; i < n; i++)
        {
            if(pass(A[i])) continue;
            if(A[i] == -1) continue;
            if(num > 0 && A[i] == -2)
            {
                num--;
                ans.push_back(i);
            }
            else if(A[i] != -2)
            {
                ans.push_back(i);
            }
        }
    }
    cout << ans.size() << endl;
    for(int i = 0; i < ans.size(); i++) cout << ans[i] + 1 << endl;
    return 0;
}




```