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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.util.stream.Collectors;

public class SolutionD extends Thread {
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                                            InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

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

    private static final FastReader scanner = new FastReader();
    private static final PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) {
        new Thread(null, new SolutionD(), "Main", 1 << 26).start();
    }

    public void run() {
        int t = scanner.nextInt();
        for (int i = 0; i < t; i++) {
            solve();
        }
        out.close();
    }

    private static final long MODULO = 1_000_000_007;

    static class Node {
        List<Node> neighbors;
        int value;
        int index;
        boolean visited = false;

        public Node(int index) {
            this.neighbors = new ArrayList<>();
            this.value = 1;
            this.index = index;
        }

        @Override
        public boolean equals(Object o) {
            if (this == o) {
                return true;
            }
            if (!(o instanceof Node)) {
                return false;
            }
            Node node = (Node) o;
            return index == node.index;
        }

        @Override
        public int hashCode() {
            return Objects.hash(index);
        }
    }

    private static void solve() {
        int n = scanner.nextInt();

        Node[] nodes = new Node[n];
        for (int i = 0; i < n; i++) {
            nodes[i] = new Node(i);
        }

        List<Node> leaves = new ArrayList<>();
        for (int i = 0; i < n - 1; i++) {
            int ui = scanner.nextInt();
            int vi = scanner.nextInt();
            nodes[ui - 1].neighbors.add(nodes[vi-1]);
            nodes[vi - 1].neighbors.add(nodes[ui-1]);
        }

        List<Long> edgesAmountVisited = new ArrayList<>();

        
        dfs(n, nodes[0], edgesAmountVisited);


        List<Long> primes = new ArrayList<>();
        int m = scanner.nextInt();
        for (int i = 0; i < m; i++) {
            long pi = scanner.nextInt();
            primes.add(pi);
        }
        for (int i = m; i < n - 1; i++) {
            primes.add(1L);
        }


        Collections.sort(primes);
        Collections.sort(edgesAmountVisited);
        while (primes.size() > n - 1) {
            primes.add((primes.remove(primes.size() - 1) * primes.remove(primes.size() - 1)) % MODULO);
        }

        long result = 0;
        for (int i = 0; i < n - 1; i++) {
            result = (result + edgesAmountVisited.get(i) * primes.get(i)) % MODULO;
        }
        System.out.println(result);
    }

    private static long dfs(int n, Node node, List<Long> edgesAmountVisited) {
        node.visited = true;

        int amountNodes = 1;
        for (Node neighbor: node.neighbors) {
            if (!neighbor.visited) {
                long amountNodesOnOtherSide = dfs(n, neighbor, edgesAmountVisited);
                amountNodes += amountNodesOnOtherSide;
                edgesAmountVisited.add(amountNodesOnOtherSide * (n - amountNodesOnOtherSide));
            }
        }
        return amountNodes;
    }

}