#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <cstdlib>
#include <unordered_map>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00000001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007 
#define REP(i,n) for(long long i = 0;i < n;++i)
#define seg_size 524288
using namespace std;
int main() {
	//0 is always 0
	vector<int> ans;
	vector<int> geko[3];
	int n;
	cin >> n;
	REP(i, n) {
		int a;
		cin >> a;
		if (a == 2) {
			ans.push_back(i + 1);
		}
		else if (a < 0) {
			geko[-a].push_back(i+1);
		}
	}
	sort(geko[1].begin(), geko[1].end());
	sort(geko[2].begin(), geko[2].end());
	for (int i = 0;i+1 < geko[2].size();i += 2) {
		ans.push_back(geko[2][i]);
		ans.push_back(geko[2][i + 1]);
	}
	if (geko[2].size() % 2 == 1) {
		if (geko[1].size() != 0) {
			ans.push_back(geko[2][geko[2].size() - 1]);
			ans.push_back(geko[1][0]);
		}
	}
	sort(ans.begin(), ans.end());
	cout << ans.size() << endl;
	for (int i = 0; i < ans.size(); ++i) {
		cout << ans[i] << endl;
	}
	return 0;
}

### Prompt

Your challenge is to write 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
#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include <map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <cstdlib>
#include <unordered_map>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00000001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007 
#define REP(i,n) for(long long i = 0;i < n;++i)
#define seg_size 524288
using namespace std;
int main() {
	//0 is always 0
	vector<int> ans;
	vector<int> geko[3];
	int n;
	cin >> n;
	REP(i, n) {
		int a;
		cin >> a;
		if (a == 2) {
			ans.push_back(i + 1);
		}
		else if (a < 0) {
			geko[-a].push_back(i+1);
		}
	}
	sort(geko[1].begin(), geko[1].end());
	sort(geko[2].begin(), geko[2].end());
	for (int i = 0;i+1 < geko[2].size();i += 2) {
		ans.push_back(geko[2][i]);
		ans.push_back(geko[2][i + 1]);
	}
	if (geko[2].size() % 2 == 1) {
		if (geko[1].size() != 0) {
			ans.push_back(geko[2][geko[2].size() - 1]);
			ans.push_back(geko[1][0]);
		}
	}
	sort(ans.begin(), ans.end());
	cout << ans.size() << endl;
	for (int i = 0; i < ans.size(); ++i) {
		cout << ans[i] << endl;
	}
	return 0;
}

```