#include "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vi v(n);
	REP(i, n)cin >> v[i];
	int p = -1;
	vi a;
	vi ans;
	REP(i, n) {
		if (v[i] == 2)ans.push_back(i);
		if (v[i] == -2)a.push_back(i);
		if (v[i] == -1 && p == -1)p = i;
	
	}
	if (a.size() % 2 == 1) {
		if (p != -1) {
			REP(i, a.size())ans.push_back(a[i]);
			ans.push_back(p);
		}
		else {
			REP(i, a.size() - 1)ans.push_back(a[i]);
		}
	}
	else {
		REP(i, a.size())ans.push_back(a[i]);
	}
	sort(ALL(ans));
	cout << ans.size() << endl;
	REP(i, ans.size()) {
		cout << ans[i]+1 << endl;
	}
	
}


### 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 "bits/stdc++.h"

#define REP(i,n) for(ll i=0;i<ll(n);++i)
#define RREP(i,n) for(ll i=ll(n)-1;i>=0;--i)
#define FOR(i,m,n) for(ll i=m;i<ll(n);++i)
#define RFOR(i,m,n) for(ll i=ll(n)-1;i>=ll(m);--i)
#define ALL(v) (v).begin(),(v).end()
#define UNIQUE(v) v.erase(unique(ALL(v)),v.end());
#define DUMP(v) REP(aa, (v).size()) { cout << v[aa]; if (aa != v.size() - 1)cout << " "; else cout << endl; }
#define INF 1000000001ll
#define MOD 1000000007ll
#define EPS 1e-9

const int dx[8] = { 1,1,0,-1,-1,-1,0,1 };
const int dy[8] = { 0,1,1,1,0,-1,-1,-1 };


using namespace std;
typedef long long ll;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
ll max(ll a, int b) { return max(a, ll(b)); }
ll max(int a, ll b) { return max(ll(a), b); }
ll min(ll a, int b) { return min(a, ll(b)); }
ll min(int a, ll b) { return min(ll(a), b); }


int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int n;
	cin >> n;
	vi v(n);
	REP(i, n)cin >> v[i];
	int p = -1;
	vi a;
	vi ans;
	REP(i, n) {
		if (v[i] == 2)ans.push_back(i);
		if (v[i] == -2)a.push_back(i);
		if (v[i] == -1 && p == -1)p = i;
	
	}
	if (a.size() % 2 == 1) {
		if (p != -1) {
			REP(i, a.size())ans.push_back(a[i]);
			ans.push_back(p);
		}
		else {
			REP(i, a.size() - 1)ans.push_back(a[i]);
		}
	}
	else {
		REP(i, a.size())ans.push_back(a[i]);
	}
	sort(ALL(ans));
	cout << ans.size() << endl;
	REP(i, ans.size()) {
		cout << ans[i]+1 << endl;
	}
	
}


```