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.*;
import java.util.*;
public class Main {
	static BufferedReader br;
	static int cin() throws Exception
	{
		return Integer.valueOf(br.readLine());
	}
	static int[] split() throws Exception
	{
		String[] cmd=br.readLine().split(" ");
		int[] ans=new int[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Integer.valueOf(cmd[i]);
		}
		return ans;
	}
	static long[] splitL() throws IOException
	{
		String[] cmd=br.readLine().split(" ");
		long[] ans=new long[cmd.length];
		for(int i=0;i<cmd.length;i++)
		{
			ans[i]=Long.valueOf(cmd[i]);
		}
		return ans;
	}
	static int gcd(int a,int b)
	{
		if(a==0)
			return b;
		if(b==0)
			return a;
		if(a>b)
			return gcd(a%b,b);
		return gcd(a,b%a);
	}
	static long mod=1000000007;
	static ArrayList<Long> count;
	static ArrayList<Integer>[] gr;
	static int[]vis;
	static long dfs(int node,int par)
	{
		if(vis[node]==1)
			return 0;
		vis[node]=1;
		long cnt=1;
		for(int i=0;i<gr[node].size();i++)
		{
			int y=gr[node].get(i);
			if(vis[y]==1)
				continue;
			long h=dfs(y,node);
			count.add(h*(gr.length-1-h));
			cnt+=h;
		}
		return cnt;
	}
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
		br=new BufferedReader(new InputStreamReader(System.in));
		BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));
		int cases=cin();
		while(cases!=0)
		{
			cases--;
			int n=cin();
			gr=new ArrayList[n+1];
			for(int i=0;i<=n;i++)
			{
				gr[i]=new ArrayList<>();
			}
			for(int i=0;i<n-1;i++)
			{
				int[]ar=split();
				int x=ar[0];
				int y=ar[1];
				gr[x].add(y);
				gr[y].add(x);
			}
			int m=cin();
			int[]p=split();
			Arrays.sort(p);
			long[] primes=new long[p.length];
			for(int i=p.length-1;i>=0;i--)
			{
				primes[p.length-1-i]=(long)p[i];
			}
			count=new ArrayList<>();
			vis=new int[n+1];
			dfs(1,-1);
			Collections.sort(count,Collections.reverseOrder());
//			System.out.println(count+" "+Arrays.toString(primes));
			if(m<n-1)
			{
				long ans=0;
				for(int i=0;i<m;i++)
				{
					ans=(ans+count.get(i)*primes[i])%mod;
				}
				for(int i=m;i<count.size();i++)
				{
					ans=(ans+count.get(i))%mod;
				}
				System.out.println(ans);
			}
			else
			{
				long pro=1;
				for(int i=0;i<=(m-n+1);i++)
				{
					pro=(pro*primes[i])%mod;
				}
				primes[m-n+1]=pro%mod;
				long ans=0;
				int ind=m-n+1;
				for(int i=0;i<count.size();i++)
				{
					ans=(ans+count.get(i)*primes[ind++])%mod;
				}
				System.out.println(ans);
			}
		}
	}
}
