1. Abridged problem statement
Given n objects, each with 1 to 4 distinct key=value properties, answer m queries of the form "how many objects contain at least these l distinct key=value pairs?" (extra properties on the object are allowed). Keys and values are short alphanumeric strings (length ≤5). n ≤ 50 000, m ≤ 100 000.

2. Detailed editorial

Overview
We need to answer many membership queries on a static set of objects. Each query asks for the count of objects that cover a given small set (size up to 4) of property constraints. A direct scan over all objects per query is too slow (n·m up to 5·10^9). Instead, we preprocess all objects by indexing counts of every possible subset of their properties.

Key observations
- Each object has at most 4 properties, so it has at most 2^4 = 16 subsets (including the empty set).
- A query is exactly one such subset; we need the total number of objects whose subsets include the query subset. Equivalently, for each object we enumerate all subsets of its properties and increment a counter for that subset representation. Then each query is a single lookup.

Preprocessing
- For object i with k properties, generate all masks from 0 to (1<<k)-1.
- For each mask, pick the corresponding subset of properties, sort them by key (to get a canonical form), serialize to a string (e.g. "key1=val1,key2=val2,").
- Maintain a hash map from this string to the count of objects that produced it.

Querying
- Read the l key=value constraints, sort them by key, serialize to the same form, look up in the map, and output the stored count (defaulting to zero if not found).

Complexities
- Preprocessing time: sum over objects of 2^ki · ki · log ki ≤ n · 16 · 4 · log 4 = O(n). Numerically ~8·10^5 operations for n=5·10^4.
- Query time: O(l · log l) per query (sorting ≤4 items) plus hash lookup, overall ~m · 4 log 4 = O(m).
- Memory: one hash map entry per distinct subset; worst-case ≤n·16 ≈ 8·10^5 entries. Should fit in the 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;
vector<vector<pair<string, string>>> objects;

vector<pair<string, string>> get_object(int cnt) {
    vector<pair<string, string>> obj;
    while(cnt--) {
        string s;
        cin >> s;
        vector<string> xy = {""};
        for(char c: s) {
            if(c == '=') {
                xy.push_back("");
                continue;
            }
            xy.back().push_back(c);
        }

        obj.push_back({xy[0], xy[1]});
    }
    return obj;
}

vector<pair<string, string>> canonize(vector<pair<string, string>> obj) {
    sort(obj.begin(), obj.end());
    return obj;
}

void read() {
    cin >> n;
    objects.resize(n);
    for(int i = 0; i < n; i++) {
        int cnt;
        cin >> cnt;
        objects[i] = canonize(get_object(cnt));
    }
}

string total_canonize(vector<pair<string, string>> obj) {
    string res;
    for(auto [x, y]: obj) {
        res += x + '=' + y + ',';
    }
    return res;
}

void solve() {
    // Each object has at most 4 properties, so it has at most 2^4 = 16 subsets
    // of properties. For every object we enumerate all subsets, canonicalize
    // each subset (sort its key=value pairs, then join with separators into a
    // single string key), and increment a hash map counting how many objects
    // contain that exact subset.
    //
    // A query lists a set of key=value constraints; an object matches iff that
    // set is one of its property subsets. So we canonicalize the query the
    // same way and answer it with one lookup into the precomputed map.

    unordered_map<string, int> cnt;
    for(int i = 0; i < n; i++) {
        for(int mask = 0; mask < (1 << objects[i].size()); mask++) {
            vector<pair<string, string>> obj;
            for(int j = 0; j < objects[i].size(); j++) {
                if(mask & (1 << j)) {
                    obj.push_back(objects[i][j]);
                }
            }
            cnt[total_canonize(obj)]++;
        }
    }

    int q;
    cin >> q;
    while(q--) {
        int c;
        cin >> c;
        vector<pair<string, string>> obj = get_object(c);
        obj = canonize(obj);
        cout << cnt[total_canonize(obj)] << '\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
from itertools import combinations

def read_ints():
    return list(map(int, sys.stdin.readline().split()))

def read_props(k):
    # read k tokens like "key=value" and split
    props = []
    for token in sys.stdin.readline().split():
        key, val = token.split('=', 1)
        props.append((key, val))
    return props

def serialize(pairs):
    # sort by key, then join into "k1=v1,k2=v2," string
    pairs_sorted = sorted(pairs)
    return ",".join(f"{k}={v}" for k,v in pairs_sorted) + ("," if pairs_sorted else "")

def main():
    input = sys.stdin.readline
    n = int(input())
    all_props = []
    for _ in range(n):
        parts = input().split()
        k = int(parts[0])
        # parse k key=val tokens
        props = [tuple(tok.split('=',1)) for tok in parts[1:1+k]]
        all_props.append(props)

    # Precompute counts of every subset
    cnt = {}
    for props in all_props:
        k = len(props)
        # for every subset size from 0..k
        for sz in range(k+1):
            # combinations from itertools
            for comb in combinations(props, sz):
                key = serialize(comb)
                cnt[key] = cnt.get(key, 0) + 1

    m = int(input())
    out = []
    for _ in range(m):
        parts = input().split()
        l = int(parts[0])
        qprops = [tuple(tok.split('=',1)) for tok in parts[1:1+l]]
        key = serialize(qprops)
        out.append(str(cnt.get(key, 0)))

    sys.stdout.write("\n".join(out))

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

5. Compressed editorial

Precompute for each object all 2^k subsets of its up-to-4 properties, store a hash map from the sorted, serialized subset to how many objects produce it. Then each query (size l ≤4) is serialized in the same way and answered by a single hash lookup.
