#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧  
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     
　　　　／　　　＼　 　  |　|     
　　　 /　　 /￣￣￣￣/　　|  
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  
　 　　　＼/＿＿＿＿/　（u　⊃  
---------------------------------------------------------------------------------------------------*/






/*
考察
a[i]=0,1は使わない
a[i]=2は全て使う
a[i]=-2はペアで使っていく
もし、a[i]=-2が奇数個でa[i]=-1があるなら、そこでペアにして使う
*/














int N, A[101];
int use[101];
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N;
    rep(i, 0, N) cin >> A[i];

    map<int, int> cnt;
    rep(i, 0, N) cnt[A[i]]++;

    // 2
    rep(i, 0, N) if (A[i] == 2) use[i] = 1;

    // -2
    int m2 = cnt[-2];
    if (m2 % 2 == 1) {
        if (0 < cnt[-1]) {
            rep(i, 0, N) if (A[i] == -1) {
                use[i] = 1;
                break;
            }
        } else m2--;
    }
    rep(i, 0, N) if (A[i] == -2 and m2) {
        use[i] = 1;
        m2--;
    }

    vector<int> ans;
    rep(i, 0, N) if (use[i]) ans.push_back(i + 1);
    
    int n = ans.size();
    printf("%d\n", n);
    rep(i, 0, n) printf("%d\n", ans[i]);
}

### 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<bits/stdc++.h>
#define rep(i,a,b) for(int i=a;i<b;i++)
#define rrep(i,a,b) for(int i=a;i>=b;i--)
#define fore(i,a) for(auto &i:a)
#define all(x) (x).begin(),(x).end()
#pragma GCC optimize ("-O3")
using namespace std; void _main(); int main() { cin.tie(0); ios::sync_with_stdio(false); _main(); }
typedef long long ll; const int inf = INT_MAX / 2; const ll infl = 1LL << 60;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a = b; return 1; } return 0; }
//---------------------------------------------------------------------------------------------------
/*---------------------------------------------------------------------------------------------------
　　　　　　　　　　　 ∧＿∧  
　　　　　 ∧＿∧ 　（´<_｀ ）　 Welcome to My Coding Space!
　　　　 （ ´_ゝ`）　/　 ⌒i     
　　　　／　　　＼　 　  |　|     
　　　 /　　 /￣￣￣￣/　　|  
　 ＿_(__ﾆつ/　    ＿/ .| .|＿＿＿＿  
　 　　　＼/＿＿＿＿/　（u　⊃  
---------------------------------------------------------------------------------------------------*/






/*
考察
a[i]=0,1は使わない
a[i]=2は全て使う
a[i]=-2はペアで使っていく
もし、a[i]=-2が奇数個でa[i]=-1があるなら、そこでペアにして使う
*/














int N, A[101];
int use[101];
//---------------------------------------------------------------------------------------------------
void _main() {
    cin >> N;
    rep(i, 0, N) cin >> A[i];

    map<int, int> cnt;
    rep(i, 0, N) cnt[A[i]]++;

    // 2
    rep(i, 0, N) if (A[i] == 2) use[i] = 1;

    // -2
    int m2 = cnt[-2];
    if (m2 % 2 == 1) {
        if (0 < cnt[-1]) {
            rep(i, 0, N) if (A[i] == -1) {
                use[i] = 1;
                break;
            }
        } else m2--;
    }
    rep(i, 0, N) if (A[i] == -2 and m2) {
        use[i] = 1;
        m2--;
    }

    vector<int> ans;
    rep(i, 0, N) if (use[i]) ans.push_back(i + 1);
    
    int n = ans.size();
    printf("%d\n", n);
    rep(i, 0, n) printf("%d\n", ans[i]);
}

```