#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using db = double;
constexpr int mod = 998244353;
constexpr int N = 1e6 + 10;

template <class T>
struct MinCostFlow {
    struct _Edge {
        int v;
        T cap, cost;
        _Edge(int v, T cap, T cost) : v(v), cap(cap), cost(cost) {}
    };
    int n;
    T inf;
    vector<_Edge> e; // e[i] 存编号为 i 的边所指向的节点 v / 流量 cap / 单位流量费用 cost
    vector<vector<int>> g; // g[u] 存储了所有以 u 为起点的边 u -> v 的编号 i
    vector<T> h; // h[u] 表示节点 u 的势函数, u -> v 的边权由 c 变成了 c + h[u] - h[v]
    vector<T> dis; // dis[t] 表示从 S 出发到达 T 的最短路径长度 (路径 单位流量 费用之和)
    vector<int> pre; // pre[v] 表示流经过了 u -> v 这条边, 记录流向 v 的这条边的编号 i
    MinCostFlow(int n) { init(n); }
    void init(int n) {
        this->n = n;
        e.clear();
        g.assign(n + 1, {});
        inf = numeric_limits<T>::max();
    }
    // 每次增广出一条 S 到 T 的最短路径, 边权为单位流量费用
    bool dij(int s, int t) {
        dis.assign(n + 1, inf);
        pre.assign(n + 1, -1);
        priority_queue<pair<T, int>, vector<pair<T, int>>, greater<pair<T, int>>> q;
        dis[s] = 0;
        q.emplace(0, s);
        while (!q.empty()) {
            auto [d, u] = q.top();
            q.pop();
            if (dis[u] != d) continue;
            for (int i : g[u]) {
                auto [v, cap, cost] = e[i];
                if (cap > 0 && dis[v] > d + h[u] - h[v] + cost) {
                    dis[v] = d + h[u] - h[v] + cost;
                    pre[v] = i;
                    q.emplace(dis[v], v);
                }
            }
        }
        return dis[t] != inf;
    }
    void add(int u, int v, T cap, T cost) {
        g[u].push_back(e.size());
        e.emplace_back(v, cap, cost);
        g[v].push_back(e.size());
        e.emplace_back(u, 0, -cost);
    }
    pair<T, T> flow(int s, int t) {
        T flow = 0;
        T cost = 0;
        h.assign(n + 1, 0);
        while (dij(s, t)) {
            for (int i = 1; i <= n; i++) h[i] += dis[i];
            // 回溯增广路径, 求路径最小容量(本次的增广流量)
            T aug = inf;
            for (int i = t; i != s; i = e[pre[i] ^ 1].v) {
                aug = min(aug, e[pre[i]].cap);
            }
            // 更新增广路径上边的容量
            for (int i = t; i != s; i = e[pre[i] ^ 1].v) {
                e[pre[i]].cap -= aug;
                e[pre[i] ^ 1].cap += aug;
            }
            flow += aug;
            cost += aug * h[t];
        }
        return make_pair(flow, cost);
    }
    struct Edge {
        int u, v;
        T cap, cost, flow;
    };
    vector<Edge> edges() {
        vector<Edge> ans;
        for (int i = 0; i < (int)e.size(); i += 2) {
            Edge x;
            x.u = e[i + 1].v;
            x.v = e[i].v;
            x.cap = e[i].cap + e[i + 1].cap;
            x.cost = e[i].cost;
            x.flow = e[i + 1].cap;
            ans.push_back(x);
        }
        return ans;
    }
};
void solve() {
    int n, m;
    cin >> n >> m;
    vector<int> l(m + 1);
    for (int i = 1; i <= m; i++) cin >> l[i];
    vector<int> col(2 * n + 1);
    vector<vector<int>> pos(m + 1);
    for (int i = 1; i <= 2 * n; i++) {
        cin >> col[i];
        pos[col[i]].push_back(i);
    }
    ll sum = 0;
    vector<int> val(2 * n + 1);
    for (int i = 1; i <= 2 * n; i++) {
        cin >> val[i];
        sum += val[i];
    }

    MinCostFlow<ll> f(2 * n + m + 3);
    int S = 2 * n + m + 2, T = 2 * n + m + 3;
    int O = 2 * n + m + 1;
    f.add(S, O, n, 0);
    for (int i = 1; i <= 2 * n; i++) f.add(O, i, 1, 0);
    for (int i = 1; i <= 2 * n; i++) {
        for (int j = i + 1; j <= 2 * n; j++) {
            f.add(i, j, 1, 0);
        }
    }
    for (int i = 1; i <= m; i++) {
        for (auto j : pos[i]) {
            f.add(j, 2 * n + i, 1, val[j]);
        }
        if (pos[i].size() < l[i]) {
            cout << "-1\n";
            return;
        }
        f.add(2 * n + i, T, pos[i].size() - l[i], 0);
    }
    auto [flow, cost] = f.flow(S, T);
    cout << (flow == n ? sum - cost : -1) << "\n";
}
int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int tt = 1;
    cin >> tt;
    while (tt--) {
        solve();
    }
    return 0;
}