1. Abridged Problem Statement
You have n integer sequences (each up to length 10). You will be given m queries; each query lists integers b₁,…,bₗ where a positive bᵢ is a required element and a negative bᵢ means "−bᵢ must NOT appear." For each query, find all sequences (in their input order) that contain every required element and contain none of the forbidden ones. Output the count of matching sequences, then print each matching sequence in the same format as input.

2. Detailed Editorial
- Constraints are tiny (n, m ≤ 10; sequence lengths ≤ 10; element values 1…100). A straightforward check per query and per sequence is efficient.
- Preprocessing: For each sequence i, store
  • A list `seqs[i]` of its elements, to reproduce output.
  • A set `present[i]` of its elements, to test membership in O(log) time.
- Query processing: For each query, read its list of signed integers. For each sequence i (in order 1…n):
  • Assume it is valid (`ok = true`).
  • For each q in the query:
    – If q > 0 and q ∉ present[i], set `ok = false`.
    – If q < 0 and (−q) ∈ present[i], set `ok = false`.
  • If `ok` remains true after checking all terms, record i as a match.
- Finally, output the number of matches, then for each matched sequence i print its length and its elements in order.

Time complexity: O(n · l) per query, n, l ≤ 10, trivial for given limits.

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<vector<int>> seqs;
vector<set<int>> present;

void read() {
    cin >> n >> m;
    seqs.assign(n, {});
    present.assign(n, {});
    for(int i = 0; i < n; i++) {
        int k;
        cin >> k;
        seqs[i].resize(k);
        cin >> seqs[i];
        for(int x: seqs[i]) {
            present[i].insert(x);
        }
    }
}

void solve() {
    // Each database sequence stores its element multiset (as a vector to
    // preserve order) and a set of present values for O(log) membership tests.
    // A query lists wanted values (positive) and unwanted values (negative);
    // a sequence matches if it contains every positive value and none of the
    // values whose negation is given. We scan sequences in input order and
    // print the matching ones in their original format.

    while(m--) {
        int l;
        cin >> l;
        vector<int> query(l);
        cin >> query;
        vector<int> matched;
        for(int i = 0; i < n; i++) {
            bool ok = true;
            for(int q: query) {
                if(q > 0 && !present[i].count(q)) {
                    ok = false;
                } else if(q < 0 && present[i].count(-q)) {
                    ok = false;
                }
            }

            if(ok) {
                matched.push_back(i);
            }
        }

        cout << matched.size() << '\n';
        for(int idx: matched) {
            cout << seqs[idx].size() << ' ' << seqs[idx] << '\n';
        }
    }
}

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

    read();
    solve();

    return 0;
}
```

4. Python Solution
```python
import sys

def main():
    data = sys.stdin.read().split()
    it = iter(data)
    n, m = map(int, (next(it), next(it)))  # number of sequences, queries

    seq = []     # list of sequences (as lists)
    has = []     # list of sets for fast membership

    # Read n sequences
    for _ in range(n):
        k = int(next(it))
        s = []
        sset = set()
        for __ in range(k):
            x = int(next(it))
            s.append(x)
            sset.add(x)
        seq.append(s)
        has.append(sset)

    # Process m queries
    out = []
    for _ in range(m):
        l = int(next(it))
        query = [int(next(it)) for __ in range(l)]

        matches = []
        for i in range(n):
            valid = True
            sset = has[i]
            # Check each term in query
            for x in query:
                if x > 0:
                    # required element
                    if x not in sset:
                        valid = False
                        break
                else:
                    # forbidden element
                    if -x in sset:
                        valid = False
                        break
            if valid:
                matches.append(i)

        # Collect output for this query
        out.append(str(len(matches)))
        for i in matches:
            s = seq[i]
            out.append(str(len(s)) + ' ' + ' '.join(map(str, s)))

    # Print all at once
    print('\n'.join(out))

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

5. Compressed Editorial
Store each sequence both as a list and as a set. For each query, test each sequence: every positive term must be in the set, every negative term's absolute value must not be. Collect and output matching sequences in input order. This runs in O(n·l) per query, more than fast enough for n, l ≤ 10.
