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

public class Main {
	
	public static void main(String[] args)
	{
		Scanner s=new Scanner(System.in);
		
		int t=s.nextInt();
		
		long z=1000000007;
		
		for(int i=0;i<t;i++)
		{
			int n=s.nextInt();
			
			ArrayList<Integer>[] arr=new ArrayList[n+1];
			
			for(int j=0;j<n+1;j++)
			{
				ArrayList<Integer> list=new ArrayList<>();
				arr[j]=list;
			}
			
			for(int j=0;j<n-1;j++)
			{
				int a=s.nextInt();
				int b=s.nextInt();
				
				arr[a].add(b);
				arr[b].add(a);
			}
			
			int m=s.nextInt();
			
			long[] prime=new long[m];
			
			for(int j=0;j<m;j++)
			{
				prime[j]=s.nextInt();
			}
			
			Arrays.sort(prime);
			
			long[] sub=new long[n+1];
			int[] visited=new int[n+1];
			
			dfs(arr,visited,1,sub);
			
			ArrayList<Long> list=new ArrayList<>();
			visited=new int[n+1];
			dfs2(arr,visited,1,sub,list,n);
			
			Collections.sort(list);
			
			long ans=0;
			
			if(m<=list.size())
			{
				int ptr1=m-1;
				int ptr2=list.size()-1;
				
				while(ptr1>=0&&ptr2>=0)
				{	
					ans=(ans%z+(prime[ptr1]%z*list.get(ptr2)%z)%z)%z;
					ptr1--;
					ptr2--;
				}
				
				while(ptr2>=0)
				{
					ans=(ans%z+list.get(ptr2)%z)%z;
					ptr2--;
				}
				
			}
			else
			{
				int ptr1=0;
				int ptr2=0;
				
				while(ptr1<m&&ptr2<list.size()-1)
				{
					ans=(ans%z+(prime[ptr1]%z*list.get(ptr2)%z)%z)%z;
					ptr1++;
					ptr2++;
				}
				
				long val=1;
				
				while(ptr1<m)
				{
					val=(val%z*prime[ptr1]%z)%z;
					ptr1++;
				}
				
				ans=(ans%z+(val%z*list.get(ptr2)%z)%z)%z;
				
			}
			
			System.out.println(ans);
			
		}
		
	}
	
	public static long dfs(ArrayList<Integer>[] arr,int[] visited,int curr,long[] sub)
	{
		visited[curr]=1;
		
		long count=0;
		
		for(int i=0;i<arr[curr].size();i++)
		{
			int now=arr[curr].get(i);
			
			if(visited[now]==0)
			{
				count=count+dfs(arr,visited,now,sub);
			}
		}
		
		sub[curr]=count+1;
		
		return sub[curr];
		
	}
	
	public static void dfs2(ArrayList<Integer>[] arr,int[] visited,int curr,long[] sub,ArrayList<Long> list,long n)
	{
		visited[curr]=1;
		
		for(int i=0;i<arr[curr].size();i++)
		{
			int now=arr[curr].get(i);
			
			if(visited[now]==0)
			{
				list.add(sub[now]*(n-sub[now]));
				dfs2(arr,visited,now,sub,list,n);
			}
		}
	}
	
}