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.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.Set;
import java.util.StringTokenizer;



public class q4 {
	static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
	static class node{
		int val;
		ArrayList<Integer> adj;
		
		public node(int n)
		{
			val = n;
			adj = new ArrayList<Integer>();
		}
	}
	public static void main(String[] args) {
		FastReader s = new FastReader();
		
		int t = s.nextInt();
		
		while(t-- > 0)
		{
			int n =s.nextInt();
			
			node[] g=  new node[n];
			for(int i=0;i<n;++i)
			{
				g[i] = new node(i);
			}
			
			for(int i=0;i<n-1;++i)
			{
				int x = s.nextInt();
				int y = s.nextInt();
				
				g[x-1].adj.add(y-1);
				g[y-1].adj.add(x-1);
			}
			int m = s.nextInt();
			
			ArrayList<Integer> pf = new ArrayList<>();
			for(int i=0;i<m;++i)
				pf.add(s.nextInt());
			
			Collections.sort(pf,Collections.reverseOrder());
			
			ArrayList<Long> a = new ArrayList<>();
			boolean[] vis = new boolean[n];
			dfs(g,0,vis,a);
			int mod = 1000000007;
			Collections.sort(a,Collections.reverseOrder());
			long ans = 0;
			if(m <= n-1)
			{
				for(int i=0;i<n-1;++i)
				{
					if(i < m)
						ans = (ans + (1L*a.get(i)*pf.get(i))%mod)%mod;
					else ans = (ans + a.get(i))%mod;
				}
				
			}
			else{
				long k = 1;
				for(int i=0;i<m - n + 2;++i)
					k = (1L*k*pf.get(i))%mod;
				ans = (1L*a.get(0)*k)%mod;
				int i = 1;
				int j = m - n + 2;
				
				while(i < n - 1)
				{
					ans = (ans + (1L*a.get(i)*pf.get(j))%mod)%mod;
					++i;++j;
				}
			}
			
			out.println(ans);
		}
		
		out.flush();
		out.close();
	}


	private static int dfs(node[] g, int curr, boolean[] vis, ArrayList<Long> a) {
		
		vis[curr] = true;
		
		ArrayList<Integer> adj = g[curr].adj;
		int k = 1;
		for(int i=0;i<adj.size();++i)
		{
			if(!vis[adj.get(i)])
				k += dfs(g,adj.get(i),vis,a);
		}
		
		a.add(1L*k*(g.length-k));
		
		return k;
	}


	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;
        }
    }

}
