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.lang.*;
import java.io.*;

public class Template {
	
	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;
  }
}
	
	static long mul(long a, long b)
	{
		return a*b%mod;
	}
	
	static class Node
	{
		ArrayList<Node> adj = new ArrayList<>();
		int subtreeSize;
		
		public void dfs(Node parent)
		{
			this.subtreeSize = 1;
			for(Node n:adj)
			{
				if(n!=parent)
				{
					n.dfs(this);
					this.subtreeSize += n.subtreeSize;
				}
			}
		}
		
	}
	
	static long mod = (long)1e9+7;

	
	    public static void main(String args[])
	    {
	    	FastReader sc = new FastReader();
	        int testcases = sc.nextInt();  // nTest is the number of treasure hunts.
	        while(testcases-- > 0)
	        {
	        	int n = sc.nextInt();
	        	Node[] nodes = new Node[n];
	        	
	        	for(int i=0;i<n;i++)
	        	{
	        		nodes[i] = new Node();
	        	}
	        	for(int i=0;i<n-1;i++)
	        	{
	        		int a= sc.nextInt()-1;
	        		int b = sc.nextInt()-1;
	        		nodes[a].adj.add(nodes[b]);
	        		nodes[b].adj.add(nodes[a]);
	        	}
	        	
	        	int primes = sc.nextInt();
	        	int []factors = new int[primes];
	        	for(int i=0;i<primes;i++)
	        	{
	        		factors[i] = sc.nextInt();
	        	}
	        	
	        	Arrays.sort(factors);
	        	n--;
	        	long[] values = new long[n];
	        	Arrays.fill(values, 1);
	        	if(primes > n)
	        	{
	        		for(int i=primes-1;i>=0;i--)
	        		{
	        			int index = Math.max(n-i-1, 0);
	        			values[index] = mul(values[index], factors[i]);
	        		}
	        	}
	        	else
	        	{
	        		
	        		for(int i=primes-1;i>=0;i--)
	        		{
	        			values[primes - i-1] = factors[i];
	        		}
	        	}
	        	
	        	long[] edgeWeights = new long[n];
	        	nodes[0].dfs(null);
	        	
	        	for(int i=1;i<=n;i++)
	        	{
	        		long times = nodes[i].subtreeSize * (long)(n +1 - nodes[i].subtreeSize);
	        		edgeWeights[i-1] = times;
	        	}
	        	
	        	Arrays.sort(edgeWeights);
	        	long ans = 0;
	        	for(int i=0;i<n;i++)
	        	{
	        		ans = (ans + mul(edgeWeights[i], values[n-1-i]))%mod; 
	        	}
	        	
	        	System.out.println(ans);
	        }
	    }
	
}