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.*;

public class Main {
	
	public static long dfs(ArrayList<ArrayList<Integer>> V, int[] o, int i, long[] br)
	{
		o[i]=1;
		long sum=1;
		for (int j=0; j<V.get(i).size(); j++)
		{
			if (o[V.get(i).get(j)]==0)
				sum+=dfs(V,o, V.get(i).get(j), br);
		}
		br[i]=sum;
		return sum;
	}
	
	public static void main (String[] args)
	{
		Scanner input=new Scanner (System.in);
		long mod=(long) (1e9+7);
		int t=input.nextInt();
		
		 
		while (t-->0)
		{
			int n=input.nextInt();
			//System.out.println(n);
			ArrayList<ArrayList<Integer> > V=new ArrayList<>();
			
			for (int i=0;i<=n;i++) V.add(new ArrayList<Integer>());
			
			for (int i=1;i<n;i++)
			{
				int x=input.nextInt();
				int y=input.nextInt();
				V.get(x).add(y);
				V.get(y).add(x);
			}
			
			
			int m=input.nextInt();
			long[] p=new long[m];
			
			for (int i=0;i<m;i++)
				p[i]=input.nextLong();
			Arrays.sort(p);
			for (int i=n-1;i<m;i++)
				p[n-2]=(p[n-2]* p[i])%mod;
			
			int[] o=new int[n+1];
			long[] br=new long[n+1];
			for (int i=0;i<=n;i++) 
			{
				o[i]=0; br[i]=0;
			}
			
			
			dfs (V, o, 1, br);
			
			for (int i=1;i<=n;i++)
				br[i]=(br[i]*((long)(n-br[i])) );
			Arrays.sort(br);
			
			int j=n-2;
			if (n-1>m) j=m-1;
			
			long ans=0;
			for (int i=n;i>=2;i--)
			{
				if (j>=0)
				{
					ans=(ans+ (br[i]*p[j])%mod)%mod;
					j--;
				}
				else ans=(ans+br[i])%mod;
			}
			System.out.println(ans);
		}
		
	}
}
