1. Abridged Problem Statement  
Given N integers a₁…aₙ (1 ≤ aᵢ ≤ M) and a modulus M, choose a subset S of indices so that the product of the selected aᵢ’s taken modulo M is as large as possible. If S is empty, the product is defined as 1. Output the maximum achievable value and any corresponding subset of lever indices (in increasing order).

2. Detailed Editorial  
We need to maximize (∏_{i∈S} aᵢ) mod M. A classic way is to use dynamic programming over prefixes and current remainder:

  • Define dp[i][r] = true if using some subset of the first i levers we can get product ≡ r (mod M).  
  • Base: dp[0][1] = true (empty set gives product 1), all other dp[0][*] = false.  
  • Transition for lever i (1-based): for each remainder r such that dp[i−1][r] is true,  
      – Skip lever i: dp[i][r] = true.  
      – Take lever i: dp[i][(r·aᵢ)%M] = true.  
  • Track a 2D array `take[i][r]` = true if the move “take lever i” was used to reach dp[i][r].  

After filling dp[N][*], scan r from M−1 down to 0 to find the maximum reachable remainder max_r. That is your answer.  

To recover which levers were taken, backtrack:
  • Start with curr = max_r at i = N.  
  • For i from N down to 1, if take[i][curr] is true, record i in the answer list and set curr to some r′ such that dp[i−1][r′] was true and (r′·aᵢ)%M == curr.  
  • Reverse the recorded list to get levers in increasing order.

Time complexity: O(N·M). With N ≤ 10⁴ and M ≤ 10³, this runs in about 10⁷ steps, which fits in 0.5 s.

3. C++ Solution
```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
}

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
}

int n, m;
vector<int> a;

void read() {
    cin >> n >> m;
    a.resize(n);
    cin >> a;
}

void solve() {
    // Subset-product maximisation modulo m via DP over reachable remainders.
    // dp[i][j] is true if some subset of the first i levers has product == j
    // (mod m), starting from the empty product 1. take[i][j] marks that the
    // best way to reach remainder j after i levers used lever i. We compute
    // the largest reachable remainder, then walk the take/dp tables backwards
    // to recover which levers were pulled, and print them in ascending order.

    vector<vector<bool>> dp(n + 1, vector<bool>(m, false));
    vector<vector<bool>> take(n + 1, vector<bool>(m, false));
    dp[0][1] = true;

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(dp[i][j]) {
                dp[i + 1][j] = true;
                dp[i + 1][(j * a[i]) % m] = true;
                take[i + 1][(j * a[i]) % m] = true;
            }
        }
    }

    int max_score = 0;
    for(int j = m - 1; j >= 0; j--) {
        if(dp[n][j]) {
            max_score = j;
            break;
        }
    }

    vector<int> levers;
    int curr = max_score;
    for(int i = n; i > 0; i--) {
        if(take[i][curr]) {
            levers.push_back(i);
            for(int j = 0; j < m; j++) {
                if(dp[i - 1][j] && (j * a[i - 1]) % m == curr) {
                    curr = j;
                    break;
                }
            }
        }
    }

    reverse(levers.begin(), levers.end());
    cout << max_score << '\n';
    cout << levers << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution with Detailed Comments  
```python
import sys
def main():
    data = sys.stdin.read().split()
    it = iter(data)
    n = int(next(it))
    m = int(next(it))
    a = [int(next(it)) for _ in range(n)]

    # dp[i][r] = can we get remainder r using first i levers?
    dp = [ [False]*m for _ in range(n+1) ]
    take = [ [False]*m for _ in range(n+1) ]

    dp[0][1] = True  # empty product = 1 mod m

    # Fill DP
    for i in range(n):
        ai = a[i] % m
        for r in range(m):
            if not dp[i][r]:
                continue
            # skip lever i
            dp[i+1][r] = True
            # take lever i
            nr = (r * ai) % m
            dp[i+1][nr] = True
            take[i+1][nr] = True

    # find best remainder
    max_r = 0
    for r in range(m-1, -1, -1):
        if dp[n][r]:
            max_r = r
            break

    # backtrack to find which indices were taken
    res = []
    curr = max_r
    for i in range(n, 0, -1):
        if not take[i][curr]:
            # lever i not used, remainder unchanged
            continue
        # lever i was used
        res.append(i)  # record index (1-based)
        # find previous remainder
        ai = a[i-1] % m
        for prev in range(m):
            if dp[i-1][prev] and (prev * ai) % m == curr:
                curr = prev
                break

    res.reverse()  # indices collected from back to front

    # output
    sys.stdout.write(str(max_r) + "\n")
    if res:
        sys.stdout.write(" ".join(map(str, res)) + "\n")

if __name__ == "__main__":
    main()
```

5. Compressed Editorial  
- DP states dp[i][r]: using first i levers, can achieve mod-product r.  
- Transitions: skip lever i → dp[i+1][r], or take lever i → dp[i+1][(r·aᵢ)%M].  
- Record “take” flags to backtrack.  
- Answer = largest r (from M−1 down) with dp[N][r] = true.  
- Backtrack by reversing take decisions, recovering indices in O(N·M).