#include <iostream>
#include <vector>
#include <map>
#include <algorithm>

using namespace std;

void solve() {
    int N;
    long long K;
    cin >> N >> K;

    vector<long long> A(N), B(N), C(N), D(N);
    for (int i = 0; i < N; i++) cin >> A[i];
    for (int i = 0; i < N; i++) cin >> B[i];
    for (int i = 0; i < N; i++) cin >> C[i];
    for (int i = 0; i < N; i++) cin >> D[i];

    // Map stores {cost: count_of_jewels_available_at_this_cost}
    map<long long, long long> pq;
    long long total_items = 0;
    long long profit = 0;

    for (int i = 0; i < N; i++) {
        // 1. Add up to B[i] buying options
        if (B[i] > 0) {
            pq[A[i]] += B[i];
            total_items += B[i];
        }

        // 2. Greedily sell up to D[i] items at price C[i]
        long long to_sell = D[i];
        long long fake_buys = 0;

        while (to_sell > 0 && !pq.empty()) {
            auto it = pq.begin();
            long long cost = it->first;
            long long count = it->second;

            // Stop if the cheapest available cost is strictly greater or equal to selling price
            if (cost >= C[i]) {
                break; 
            }

            long long take = min(to_sell, count);
            profit += take * (C[i] - cost);
            fake_buys += take;
            to_sell -= take;

            it->second -= take;
            if (it->second == 0) {
                pq.erase(it);
            }
        }

        // Add back fake buys to allow future upgrading of the sales prices
        if (fake_buys > 0) {
            pq[C[i]] += fake_buys;
            // Notice total_items doesn't increase here because the fake_buys 
            // directly replace the actual buys we pulled out above.
        }

        // 3. Ensure we don't violate the storage capacity constraint K
        while (total_items > K) {
            auto it = prev(pq.end()); // Look at the elements with the highest cost
            long long count = it->second;

            long long remove_cnt = min(total_items - K, count);
            total_items -= remove_cnt;
            it->second -= remove_cnt;
            
            if (it->second == 0) {
                pq.erase(it);
            }
        }
    }

    cout << profit << "\n";
}

int main() {
    // Fast I/O
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int T;
    if (cin >> T) {
        while (T--) {
            solve();
        }
    }
    return 0;
}