#include <bits/stdc++.h>

#define x first
#define y second
#define pb push_back
#define int long long 

using namespace std;
typedef long long LL;
typedef pair<int,int> pii;
typedef pair<double, double> PDD;
 
const int N = 1E4 + 5, M = 2E6 + 10, INF = 1e10;
const int MOD = 1e9 + 7;
const LL LINF = 2e18;
const double eps = 1e-9;

int n, m, S, T;
int h[N], e[M], ne[M], w[M], f[M], idx;
int hd[N];//预处理建立虚拟远点到所有的点的距离
int d[N], pre[N], incf[N], q[N];
bool st[N];

void add(int a, int b, int c, int d){
    e[idx] = b, f[idx] = c, w[idx] =  d, ne[idx] = h[a], h[a] = idx ++;
    e[idx] = a, f[idx] = 0, w[idx] = -d, ne[idx] = h[b], h[b] = idx ++;

}


bool spfa()
{
    int hh = 0, tt = 1;
    memset(d, 0x3f, sizeof d);
    memset(incf, 0, sizeof incf);
    q[0] = S, d[S] = 0, incf[S] = INF;
    while (hh != tt)
    {
        int t = q[hh ++ ];
        if (hh == N) hh = 0;
        st[t] = false;

        for (int i = h[t]; ~i; i = ne[i])
        {
            int ver = e[i];
            if (f[i] && d[ver] > d[t] + w[i])
            {
                d[ver] = d[t] + w[i];
                pre[ver] = i;
                incf[ver] = min(incf[t], f[i]);
                if (!st[ver])
                {
                    q[tt ++ ] = ver;
                    if (tt == N) tt = 0;
                    st[ver] = true;
                }
            }
        }
    }
    return incf[T] > 0;
}

void EK(int& flow, int& cost){
    flow = cost = 0;
    while(spfa()){
        int t = incf[T];
        //t * d[T]，每一次是整条增广路径上的边流量都增加t
        //*d[T] 直接乘上费用
        flow += t, cost += t * d[T]; 
        
        for(int i = T; i != S; i = e[pre[i] ^ 1]){
            f[pre[i]] -= t;
            f[pre[i] ^ 1] += t;
        }
    }
}


int c[N], v[N], l[N];

void solve(){
	cin >> n >> m;
	S = 0, T = n * 2 + m + 1;

	idx = 0;
	memset(h, -1, sizeof h);
	LL mi = 0, ans = 0, sum = 0;
	for (int i = 1; i <= m; i ++){
		cin >> l[i];
		sum += l[i];
	}
	for (int i = 1; i <= n * 2; i ++)cin >> c[i];
	for (int i = 1; i <= n * 2; i ++)cin >> v[i];
	
	if(sum > n){
		cout << "-1" << '\n';
		return;
	}


	for (int i = 1; i + 1 <= n * 2; i ++){
		add(i, i + 1, (i + 1) / 2, - INF);
		add(i, i + 1, n, 0);
		mi += (i + 1) / 2 * (-INF);
		
	}

	add(2 * n, T, n, 0);
	for (int i = 1; i <= m; i ++){
		add(S, 2 * n + i, l[i], -INF);
		add(S, 2 * n + i, n, 0);
		mi += l[i] * (-INF);
	}

	for (int i = 1; i <= 2 * n; i ++)
		add(2 * n + c[i], i, 1, -v[i]);
	
	int flow, cost;
	EK(flow, cost);
	if(cost > mi) cout << "-1" << '\n';
	else cout << -(cost - mi) << '\n';
}

signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t = 1;cin >>t;
	while(t --) solve();
	return 0;
}