You are given a tree that consists of n nodes. You should label each of its n-1 edges with an integer in such way that satisfies the following conditions: 

  * each integer must be greater than 0; 
  * the product of all n-1 numbers should be equal to k; 
  * the number of 1-s among all n-1 integers must be minimum possible. 



Let's define f(u,v) as the sum of the numbers on the simple path from node u to node v. Also, let ∑_{i=1}^{n-1} ∑_{j=i+1}^n f(i,j) be a distribution index of the tree.

Find the maximum possible distribution index you can get. Since answer can be too large, print it modulo 10^9 + 7.

In this problem, since the number k can be large, the result of the prime factorization of k is given instead.

Input

The first line contains one integer t (1 ≤ t ≤ 100) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 10^5) — the number of nodes in the tree.

Each of the next n-1 lines describes an edge: the i-th line contains two integers u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) — indices of vertices connected by the i-th edge.

Next line contains a single integer m (1 ≤ m ≤ 6 ⋅ 10^4) — the number of prime factors of k.

Next line contains m prime numbers p_1, p_2, …, p_m (2 ≤ p_i < 6 ⋅ 10^4) such that k = p_1 ⋅ p_2 ⋅ … ⋅ p_m.

It is guaranteed that the sum of n over all test cases doesn't exceed 10^5, the sum of m over all test cases doesn't exceed 6 ⋅ 10^4, and the given edges for each test cases form a tree.

Output

Print the maximum distribution index you can get. Since answer can be too large, print it modulo 10^9+7.

Example

Input


3
4
1 2
2 3
3 4
2
2 2
4
3 4
1 3
3 2
2
3 2
7
6 1
2 3
4 6
7 3
5 1
3 6
4
7 5 13 3


Output


17
18
286

Note

In the first test case, one of the optimal ways is on the following image:

<image>

In this case, f(1,2)=1, f(1,3)=3, f(1,4)=5, f(2,3)=2, f(2,4)=4, f(3,4)=2, so the sum of these 6 numbers is 17.

In the second test case, one of the optimal ways is on the following image:

<image>

In this case, f(1,2)=3, f(1,3)=1, f(1,4)=4, f(2,3)=2, f(2,4)=5, f(3,4)=3, so the sum of these 6 numbers is 18.
import java.io.*;
import java.util.*;

/*
polyakoff
*/

public class Main {

    static FastReader in;
    static PrintWriter out;
    static Random rand = new Random();
    static final int oo = (int) 2e9;
    static final long OO = (long) 4e18;
    static final int MOD = (int) 1e9 + 7;

    static int n, m;
    static ArrayList<Integer>[] g;
    static ArrayList<Edge> edges;
    static long[] p;
    static int[] size;

    static class Edge {
        int u, v, ind;

        Edge(int u, int v, int ind) {
            this.u = u;
            this.v = v;
            this.ind = ind;
        }
    }

    static void dfs(int u) {
        size[u] = 1;
        for (int v : g[u]) {
            if (size[v] == 0) {
                dfs(v);
                size[u] += size[v];
            }
        }
    }

    static void solve() {
        n = in.nextInt();

        g = new ArrayList[n]; Arrays.setAll(g, i -> new ArrayList<>());
        edges = new ArrayList<>();
        for (int i = 0; i < n - 1; i++) {
            int u = in.nextInt() - 1;
            int v = in.nextInt() - 1;
            g[u].add(v);
            g[v].add(u);
            edges.add(new Edge(u, v, i));
        }

        m = in.nextInt();

        long[] p2 = new long[m];
        for (int i = 0; i < m; i++)
            p2[i] = in.nextInt();
        Arrays.sort(p2);

        p = new long[n - 1];
        if (m > n - 1) {
            long fi = 1;
            for (int i = n - 2; i < m; i++)
                fi = (fi * p2[i]) % MOD;
            p[0] = fi;
            for (int i = 1; i < n - 1; i++)
                p[i] = p2[n - 2 - i];
        } else {
            for (int i = 0; i < m; i++)
                p[i] = p2[m - 1 - i];
            for (int i = m; i < n - 1; i++)
                p[i] = 1;
        }

//        dbg(Arrays.toString(p), '\n');

        size = new int[n];
        dfs(0);

//        dbg(Arrays.toString(size), '\n');

        Collections.sort(edges, (e1, e2) -> {
            int min1 = Math.min(size[e1.u], size[e1.v]);
            int min2 = Math.min(size[e2.u], size[e2.v]);
            return -Long.compare((long) min1 * (n - min1), (long) min2 * (n - min2));
        });

        long ans = 0;
        for (int i = 0; i < n - 1; i++) {
            Edge e = edges.get(i);
            int min = Math.min(size[e.u], size[e.v]);
            ans = (ans + p[i] * min * (n - min)) % MOD;
        }

        out.println(ans);




    }

    public static void main(String[] args) throws FileNotFoundException {
        in = new FastReader(System.in);
//        in = new FastReader(new FileInputStream("sometext.txt"));
        out = new PrintWriter(System.out);
//        out = new PrintWriter(new FileOutputStream("output.txt"));

        int T = 1;
        T = in.nextInt();
        while (T-- > 0)
            solve();

//        Thread thread = new Thread(null, () -> {
//            solve();
//        }, "Main", 1 << 28);
//        thread.start();
//        try {
//            thread.join();
//        } catch (InterruptedException e) {
//            e.printStackTrace();
//        }

//        out.flush();
        out.close();
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        FastReader(InputStream is) {
            br = new BufferedReader(new InputStreamReader(is));
        }

        Integer nextInt() {
            return Integer.parseInt(next());
        }

        Long nextLong() {
            return Long.parseLong(next());
        }

        Double nextDouble() {
            return Double.parseDouble(next());
        }

        String next() {
            while (st == null || !st.hasMoreTokens()) {
                st = new StringTokenizer(nextLine());
            }
            return st.nextToken();
        }

        String nextLine() {
            String s = "";
            try {
                s = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return s;
        }
    }

    static void dbg(Object... args) {
        for (Object arg : args)
            out.print(" " + arg);
        out.flush();
    }

}