#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <functional>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cassert>
#include <exception>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<P> vp;
#define rep(i,a,n) for(ll i = (a);i < (n);i++)
#define per(i,a,n) for(ll i = (a);i > (n);i--)
#define lep(i,a,n) for(ll i = (a);i <= (n);i++)
#define pel(i,a,n) for(ll i = (a);i >= (n);i--)
#define clr(a,b) memset((a),(b),sizeof(a))
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define sz size()
#define print(X) cout << (X) << endl
const ll INF = 1e+9+7;
ll n,m,l;
string s,t;
ll d[200010],dp[550][550];
ll box[200010];

int main(){
  cin >> n;
  map<int,int> test;
  rep(i,0,n)cin >> d[i],test[d[i]]++;

  map<int,int> memo;
  if(test[2] > 0)memo[2] = test[2];
  if(test[-2] > 0){
    if(!(test[-2] % 2)){
      memo[-2] = test[-2];
    }
    else{
      if(test[-1]){
        memo[-2] = test[-2];
        memo[-1] = 1;
      }else{
        memo[-2] = test[-2] - 1;
      }
    }
  }
  // no 2 & -2
  //if(memo.empty())if(test[1] > 0)memo[1] = 1;
  //if(memo.empty())if(test[-1] > 1)memo[-1] = 2;
  // ans
  vi vec;
  ll ans = 1;
  rep(i,0,n){
    if(memo[d[i]] > 0){
      //cout << i + 1 << ",d = " << d[i] << endl;
      vec.pb(i + 1);
      memo[d[i]]--;
      ans *= d[i];
    }
  }
  // cout << "ans:" << ans << endl;
  print(vec.sz);
  rep(i,0,vec.sz)print(vec[i]);
  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 <cstring>
#include <cstdlib>
#include <cmath>
#include <algorithm>
#include <functional>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <bitset>
#include <cassert>
#include <exception>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,ll> P;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<string> vs;
typedef vector<P> vp;
#define rep(i,a,n) for(ll i = (a);i < (n);i++)
#define per(i,a,n) for(ll i = (a);i > (n);i--)
#define lep(i,a,n) for(ll i = (a);i <= (n);i++)
#define pel(i,a,n) for(ll i = (a);i >= (n);i--)
#define clr(a,b) memset((a),(b),sizeof(a))
#define pb push_back
#define mp make_pair
#define all(c) (c).begin(),(c).end()
#define sz size()
#define print(X) cout << (X) << endl
const ll INF = 1e+9+7;
ll n,m,l;
string s,t;
ll d[200010],dp[550][550];
ll box[200010];

int main(){
  cin >> n;
  map<int,int> test;
  rep(i,0,n)cin >> d[i],test[d[i]]++;

  map<int,int> memo;
  if(test[2] > 0)memo[2] = test[2];
  if(test[-2] > 0){
    if(!(test[-2] % 2)){
      memo[-2] = test[-2];
    }
    else{
      if(test[-1]){
        memo[-2] = test[-2];
        memo[-1] = 1;
      }else{
        memo[-2] = test[-2] - 1;
      }
    }
  }
  // no 2 & -2
  //if(memo.empty())if(test[1] > 0)memo[1] = 1;
  //if(memo.empty())if(test[-1] > 1)memo[-1] = 2;
  // ans
  vi vec;
  ll ans = 1;
  rep(i,0,n){
    if(memo[d[i]] > 0){
      //cout << i + 1 << ",d = " << d[i] << endl;
      vec.pb(i + 1);
      memo[d[i]]--;
      ans *= d[i];
    }
  }
  // cout << "ans:" << ans << endl;
  print(vec.sz);
  rep(i,0,vec.sz)print(vec[i]);
  return 0;
}

```