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.util.*;
import java.util.Map.Entry;
import java.math.*;
import java.io.*;

public class Main {
	final static int MOD = 1000000007;
	static List<Integer>[] g;
	static int[] sum;
	static List<Long> list;
	static int n;

	public static void main(String[] args) throws FileNotFoundException {
		// InputReader in = new InputReader(System.in);
		// Scanner in = new Scanner(System.in);
		Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
		PrintWriter out = new PrintWriter(System.out);
		// InputReader in = new InputReader(new
		// File("ethan_traverses_a_tree.txt"));
		// PrintWriter out = new PrintWriter(new
		// File("ethan_traverses_a_tree-output.txt"));

		int pi = in.nextInt();
		for (int qi = 0; qi < pi; qi++) {
			n = in.nextInt();
			g = new ArrayList[n];
			sum = new int[n];
			list = new ArrayList<Long>();

			for (int i = 0; i < n; i++) {
				g[i] = new ArrayList<Integer>();
			}

			for (int i = 0; i < n - 1; i++) {
				int from = in.nextInt() - 1;
				int to = in.nextInt() - 1;

				g[from].add(to);
				g[to].add(from);
			}

			int m = in.nextInt();
			List<Long> klist = new ArrayList<Long>();
			for (int i = 0; i < m; i++) {
				klist.add(in.nextLong());
			}
			for (int i = 0; i < n - 1 - m; i++) {
				klist.add((long) 1);
			}
			Collections.sort(klist);
			while (klist.size() > n - 1) {
				int s = klist.size();
				long k1 = klist.get(s - 1);
				long k2 = klist.get(s - 2);

				klist.remove(s - 1);
				klist.remove(s - 2);
				klist.add(k1 * k2 % MOD);
			}

			getsum(0, -1);

			Collections.sort(list);

			long ans = 0;
			for (int i = 0; i < n - 1; i++) {
				long k1 = list.get(i) % MOD;
				long k2 = klist.get(i) % MOD;
				ans = (ans + (k1 * k2) % MOD) % MOD;
			}
			out.printf("%d\n", ans);
		}

		out.close();
	}

	static public void getsum(int now, int pre) {
		int nowsum = 1;

		for (Integer to : g[now]) {
			if (to != pre) {
				getsum(to, now);
				nowsum += sum[to];
				list.add((long) sum[to] * (n - sum[to]));
			}
		}

		sum[now] = nowsum;
	}

	static class InputReader {
		BufferedReader br;
		StringTokenizer st;

		public InputReader(File f) {
			try {
				br = new BufferedReader(new FileReader(f));
			} catch (FileNotFoundException e) {
				e.printStackTrace();
			}
		}

		public InputReader(InputStream in) {
			br = new BufferedReader(new InputStreamReader(in));
		}

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

		public boolean hasNext() {
			while (st == null || !st.hasMoreTokens()) {
				String s = null;
				try {
					s = br.readLine();
				} catch (IOException e) {
					e.printStackTrace();
				}
				if (s == null)
					return false;
				st = new StringTokenizer(s);
			}
			return true;
		}

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

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

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