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 `seq[i]` of its elements, to reproduce output.  
  • A hash‐set `has[i]` of its elements, to test membership in O(1).  
- Query processing: For each query, read its list L of signed integers. For each sequence i (in order 1…n):  
  • Assume it is valid (`ok = true`).  
  • For each x in L:  
    – If x > 0 and x ∉ has[i], set `ok = false`.  
    – If x < 0 and (−x) ∈ has[i], set `ok = false`.  
  • If `ok` remains true after checking all x, 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 with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int n, m;
    cin >> n >> m;                    // Read number of sequences and number of queries

    vector<vector<int>> seq(n);       // seq[i]: the i-th sequence's elements
    vector<unordered_set<int>> has(n);  // has[i]: set of elements in the i-th sequence

    // Read all sequences
    for (int i = 0; i < n; i++) {
        int k;
        cin >> k;                     // length of sequence i
        seq[i].resize(k);
        for (int j = 0; j < k; j++) {
            int x;
            cin >> x;                 // read element
            seq[i][j] = x;            // store in list form
            has[i].insert(x);         // store in set for O(1) membership test
        }
    }

    // Process each query
    while (m--) {
        int l;
        cin >> l;                     // number of query terms
        vector<int> query(l);
        for (int i = 0; i < l; i++) {
            cin >> query[i];          // read each signed integer
        }

        vector<int> matches;          // store indices of sequences that match

        // Check each sequence against query
        for (int i = 0; i < n; i++) {
            bool ok = true;           // assume sequence i matches unless proven otherwise

            // Test all query terms
            for (int x : query) {
                if (x > 0) {
                    // x is required: if not in the sequence, reject it
                    if (!has[i].count(x)) {
                        ok = false;
                        break;
                    }
                } else {
                    // x is negative: -x must NOT be present
                    if (has[i].count(-x)) {
                        ok = false;
                        break;
                    }
                }
            }
            if (ok) {
                matches.push_back(i); // record matching sequence index
            }
        }

        // Output results for this query
        cout << matches.size() << '\n';
        for (int idx : matches) {
            // print sequence in original format: length followed by elements
            cout << seq[idx].size();
            for (int v : seq[idx]) {
                cout << ' ' << v;
            }
            cout << '\n';
        }
    }

    return 0;
}
```

4. Python Solution with Detailed Comments  
```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.