#include<iostream>
#include<string>
#include<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(iinclude<snt i=sta;i<=n;i++)
const ld eps = 1e-6;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;

int n;
void solve() {
	cin >> n;
	vector<int> a(n);
	rep(i, n)cin >> a[i];
	int chk = -1;
	rep(i, n) {
		if (a[i] == -1) {
			chk = i; break;
		}
	}
	if (chk >= 0) {

		vector<int> ans;
		vector<int> re;
		rep(i, n) {
			if (a[i] == 2) {
				ans.push_back(i);
			}
			if (a[i] == -2) {
				re.push_back(i);
			}
		}
		if (re.size() % 2)ans.push_back(chk);
		rep(i, re.size())ans.push_back(re[i]);
		sort(ans.begin(), ans.end());

		cout << ans.size() << endl;
		rep(i, ans.size()) {
			cout << ans[i]+1 << endl;
		}
	}
	else {
		vector<int> ans;
		vector<int> re;
		rep(i, n) {
			if (a[i] == 2) {
				ans.push_back(i);
			}
			if (a[i] == -2) {
				re.push_back(i);
			}
		}
		if (re.size() % 2)re.pop_back();
		rep(i, re.size())ans.push_back(re[i]);
		sort(ans.begin(), ans.end());
		cout << ans.size() << endl;
		rep(i, ans.size()) {
			cout << ans[i]+1 << endl;
		}
		
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	//stop
		return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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<cstdio>
#include<vector>
#include<cmath>
#include<algorithm>
#include<functional>
#include<iomanip>
#include<queue>
#include<ciso646>
#include<random>
#include<map>
#include<set>
#include<complex>
#include<bitset>
#include<stack>
#include<unordered_map>
#include<utility>
using namespace std;
typedef long long ll;
typedef unsigned long long ul;
typedef unsigned int ui;
const ll mod = 1000000007;
typedef double ld;
typedef complex<ld> Point;
const ll INF = mod * mod;
typedef pair<int, int> P;
#define stop char nyaa;cin>>nyaa;
#define rep(i,n) for(int i=0;i<n;i++)
#define per(i,n) for(int i=n-1;i>=0;i--)
#define Rep(i,sta,n) for(int i=sta;i<n;i++)
#define rep1(i,n) for(int i=1;i<=n;i++)
#define per1(i,n) for(int i=n;i>=1;i--)
#define Rep1(i,sta,n) for(iinclude<snt i=sta;i<=n;i++)
const ld eps = 1e-6;
const ld pi = acos(-1.0);
typedef pair<ld, ld> LDP;
typedef pair<ll, ll> LP;

int n;
void solve() {
	cin >> n;
	vector<int> a(n);
	rep(i, n)cin >> a[i];
	int chk = -1;
	rep(i, n) {
		if (a[i] == -1) {
			chk = i; break;
		}
	}
	if (chk >= 0) {

		vector<int> ans;
		vector<int> re;
		rep(i, n) {
			if (a[i] == 2) {
				ans.push_back(i);
			}
			if (a[i] == -2) {
				re.push_back(i);
			}
		}
		if (re.size() % 2)ans.push_back(chk);
		rep(i, re.size())ans.push_back(re[i]);
		sort(ans.begin(), ans.end());

		cout << ans.size() << endl;
		rep(i, ans.size()) {
			cout << ans[i]+1 << endl;
		}
	}
	else {
		vector<int> ans;
		vector<int> re;
		rep(i, n) {
			if (a[i] == 2) {
				ans.push_back(i);
			}
			if (a[i] == -2) {
				re.push_back(i);
			}
		}
		if (re.size() % 2)re.pop_back();
		rep(i, re.size())ans.push_back(re[i]);
		sort(ans.begin(), ans.end());
		cout << ans.size() << endl;
		rep(i, ans.size()) {
			cout << ans[i]+1 << endl;
		}
		
	}
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	solve();
	//stop
		return 0;
}

```