#include <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <array>
#include <bitset>
#include <assert.h>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

int main() {
    int n = in();
    VI a(n);
    REP(i,n) a[i] = in();

    int m2 = 0, m1 = 0;
    REP(i,n){
        if (a[i] == -2) m2++;
        if (a[i] == -1) m1++;
    }

    VI ans;
    if (m2 % 2 == 0){
        REP(i,n){
            if (abs(a[i]) == 2) ans.push_back(i);
        }
    }else{
        if (m1){
            int c = 1;
            REP(i,n){
                if (abs(a[i]) == 2 || (a[i] == -1 && c == 1)){
                    ans.push_back(i);
                    if (a[i] == -1) c = 0;
                }
            }
        }else{
            int c = m2 - 1;
            REP(i,n){
                if (a[i] == 2 || (a[i] == -2 && c > 0)){
                    ans.push_back(i);
                    if (a[i] == -2) c--;
                }
            }
        }
    }

    cout << ans.size() << endl;
    for (int x : ans) cout << x + 1 << endl;

    return 0;
}

### Prompt

Generate 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 <iostream>
#include <fstream>
#include <cstdio>
#include <cmath>
#include <vector>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <algorithm>
#include <array>
#include <bitset>
#include <assert.h>
using namespace std;

#define REP(i,n) for(int i=0; i<n; ++i)
#define FOR(i,a,b) for(int i=a; i<=b; ++i)
#define FORR(i,a,b) for (int i=a; i>=b; --i)
#define ALL(c) (c).begin(), (c).end()

typedef long long ll;
typedef vector<int> VI;
typedef vector<ll> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> P;
typedef pair<ll,ll> PL;

int in() { int x; scanf("%d", &x); return x; }
ll lin() { ll x; scanf("%lld", &x); return x; }

int main() {
    int n = in();
    VI a(n);
    REP(i,n) a[i] = in();

    int m2 = 0, m1 = 0;
    REP(i,n){
        if (a[i] == -2) m2++;
        if (a[i] == -1) m1++;
    }

    VI ans;
    if (m2 % 2 == 0){
        REP(i,n){
            if (abs(a[i]) == 2) ans.push_back(i);
        }
    }else{
        if (m1){
            int c = 1;
            REP(i,n){
                if (abs(a[i]) == 2 || (a[i] == -1 && c == 1)){
                    ans.push_back(i);
                    if (a[i] == -1) c = 0;
                }
            }
        }else{
            int c = m2 - 1;
            REP(i,n){
                if (a[i] == 2 || (a[i] == -2 && c > 0)){
                    ans.push_back(i);
                    if (a[i] == -2) c--;
                }
            }
        }
    }

    cout << ans.size() << endl;
    for (int x : ans) cout << x + 1 << endl;

    return 0;
}

```