#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < int(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = int(n) - 1; (i) >= int(m); -- (i))
#define ALL(x) begin(x), end(x)
#define dump(x) cerr << #x " = " << x << endl
#define unittest_name_helper(counter) unittest_ ## counter
#define unittest_name(counter) unittest_name_helper(counter)
#define unittest __attribute__((constructor)) void unittest_name(__COUNTER__) ()
using ll = long long;
using namespace std;

vector<int> realize(int n, vector<int> const & a, vector<bool> const & used) {
    vector<int> s;
    REP (i, n) if (used[i]) {
        s.push_back(i);
    }
    return s;
}

vector<int> solve(int n, vector<int> const & a) {
    vector<bool> used(n);
    int minus_two = 0;
    bool minus_one = false;
    REP (i, n) {
        switch (a[i]) {
            case  2: used[i] = true; break;  // use 2
            case  1: /* nop */ break;
            case  0: /* nop */ break;
            case -1: minus_one = true; break;
            case -2: minus_two += 1; break;
            default: assert (false);
        }
    }
    if (minus_two % 2 == 0) {
        minus_one = false;
    } else {
        if (not minus_one) {
            minus_two -= 1;
        }
    }
    // use -2
    int i = 0;
    while (minus_two --) {
        while (a[i] != -2) ++ i;
        used[i] = true;
        ++ i;
    }
    // use -1
    if (minus_one) {
        vector<bool> nused;
        REP (i, n) if (a[i] == -1) {
            if (nused.empty() or realize(n, a, used) < realize(n, a, nused)) {
                nused = used;
                nused[i] = true;
            }
        }
        used = nused;
    }
    return realize(n, a, used);
}

int main() {
    int n; scanf("%d", &n);
    vector<int> a(n); REP (i, n) scanf("%d", &a[i]);
    vector<int> result = solve(n, a);
    printf("%d\n", int(result.size()));
    for (int i : result) printf("%d\n", i + 1);
    return 0;
}


### Prompt

Please create a solution in CPP 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 (int i = 0; (i) < int(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < int(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
#define REP3R(i, m, n) for (int i = int(n) - 1; (i) >= int(m); -- (i))
#define ALL(x) begin(x), end(x)
#define dump(x) cerr << #x " = " << x << endl
#define unittest_name_helper(counter) unittest_ ## counter
#define unittest_name(counter) unittest_name_helper(counter)
#define unittest __attribute__((constructor)) void unittest_name(__COUNTER__) ()
using ll = long long;
using namespace std;

vector<int> realize(int n, vector<int> const & a, vector<bool> const & used) {
    vector<int> s;
    REP (i, n) if (used[i]) {
        s.push_back(i);
    }
    return s;
}

vector<int> solve(int n, vector<int> const & a) {
    vector<bool> used(n);
    int minus_two = 0;
    bool minus_one = false;
    REP (i, n) {
        switch (a[i]) {
            case  2: used[i] = true; break;  // use 2
            case  1: /* nop */ break;
            case  0: /* nop */ break;
            case -1: minus_one = true; break;
            case -2: minus_two += 1; break;
            default: assert (false);
        }
    }
    if (minus_two % 2 == 0) {
        minus_one = false;
    } else {
        if (not minus_one) {
            minus_two -= 1;
        }
    }
    // use -2
    int i = 0;
    while (minus_two --) {
        while (a[i] != -2) ++ i;
        used[i] = true;
        ++ i;
    }
    // use -1
    if (minus_one) {
        vector<bool> nused;
        REP (i, n) if (a[i] == -1) {
            if (nused.empty() or realize(n, a, used) < realize(n, a, nused)) {
                nused = used;
                nused[i] = true;
            }
        }
        used = nused;
    }
    return realize(n, a, used);
}

int main() {
    int n; scanf("%d", &n);
    vector<int> a(n); REP (i, n) scanf("%d", &a[i]);
    vector<int> result = solve(n, a);
    printf("%d\n", int(result.size()));
    for (int i : result) printf("%d\n", i + 1);
    return 0;
}


```