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.io.*;
public class R665D{
	static HashMap<Integer,HashSet<Integer>> edges;
	static int[] des;
	public static void main(String[] main) throws Exception{
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		StringTokenizer st = new StringTokenizer(br.readLine());
		int t = Integer.parseInt(st.nextToken());
		int mod = 1000000007;
		for(int trial = 1; trial <= t; trial++){
			st = new StringTokenizer(br.readLine());
			int n = Integer.parseInt(st.nextToken());
			edges = new HashMap<Integer,HashSet<Integer>>();
			for(int i = 0; i < n; i++)
				edges.put(i, new HashSet<Integer>());
			for(int i = 0; i < n-1; i++) {
				st = new StringTokenizer(br.readLine());
				int u = Integer.parseInt(st.nextToken())-1;
				int v = Integer.parseInt(st.nextToken())-1;
				edges.get(u).add(v);
				edges.get(v).add(u);
			}
			st = new StringTokenizer(br.readLine());
			int m = Integer.parseInt(st.nextToken());
			st = new StringTokenizer(br.readLine());
			ArrayList<Integer> primes = new ArrayList<Integer>();
			for(int i = 0; i < m; i++)
				primes.add(Integer.parseInt(st.nextToken()));
			for(int i = 0; i < n-1-m; i++)
				primes.add(1);
			Collections.sort(primes);
			if(m > n-1) {
				long last = 1;
				for(int i = n-2; i < m; i++)
					last = (last*primes.get(i))%mod;
				primes.add(n-2, (int)last);
			}
			des = new int[n];
			dfs(0, -1);
			ArrayList<Long> f = new ArrayList<Long>();
			for(int i = 1; i < n; i++) {
				f.add((long)des[i]*(n-des[i]));
			}
			Collections.sort(f);
			long ans = 0;
			int index = 0;
			for(long i: f) {
				ans = (ans + i*primes.get(index))%mod;
				index++;
			}
			out.println(ans);
		}
		out.close();
	}
	public static void dfs(int cur, int par) {
		for(int i: edges.get(cur)) {
			if(i != par) {
				dfs(i, cur);
				des[cur] += des[i];
			}
		}
		des[cur]++;
	}
}