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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;


public class D
{
	static int subtreesize[];
	static ArrayList<Long> ans;
	static ArrayList<Integer> graph[];static int i;
	static int mod = (int)((1e9)+7);
	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);
		int t = Integer.parseInt(br.readLine());
		while(t-->0)
		{
			int n =Integer.parseInt(br.readLine());
			graph = new ArrayList[n];
			for(int i=0;i<n;i++)
			{
				graph[i] = new ArrayList<Integer>();
			}
			subtreesize = new int[n];
			for(int i=0;i<n-1;i++)
			{
				String[] inp = br.readLine().split(" ");
				int u = Integer.parseInt(inp[0]);
				int v = Integer.parseInt(inp[1]);
				graph[u-1].add(v-1);
				graph[v-1].add(u-1);
			}
			ans = new ArrayList<Long>();
			dfs(0,-1);
			int m = Integer.parseInt(br.readLine());
			PriorityQueue<Long>factorsLarge = new PriorityQueue<Long>(Collections.reverseOrder());
			String[] inp = br.readLine().split(" ");
			for(int i=0;i<m;i++) {
				factorsLarge.add(Long.parseLong(inp[i]));
			}
			for(int i=m;i<n-1;i++)factorsLarge.add(1l);
			long max = factorsLarge.poll();
			for(int i=n-1;i<m;i++)
			{
				max *=factorsLarge.poll();
				max%=mod;		
			}
			Collections.sort(ans);
			Collections.reverse(ans);
			long answer = max*ans.get(0);
			for(int i=1;i<ans.size();i++)
			{
				if(!factorsLarge.isEmpty())
					answer+=(factorsLarge.poll()*ans.get(i))%mod;
				else
					answer+= ans.get(i)%mod;
				answer%=mod;
			}

			pw.println(answer);
		}
		
		pw.flush();
		pw.close();
	}
	
	static void dfs (int node, int parent)
	{
		subtreesize[node]=1;
		for(int u : graph[node])
		{
			if(u!=parent) {
			dfs(u,node);
			subtreesize[node] +=subtreesize[u];
			}
		}
		if(parent!=-1)
			ans.add(subtreesize[node]*1l*(graph.length-subtreesize[node]));
	}

}
