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.

// Problem: D. Maximum Distributed Tree
// Contest: Codeforces - Codeforces Round #665 (Div. 2)
// URL: https://codeforces.com/contest/1401/problem/D
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// Powered by CP Editor (https://github.com/cpeditor/cpeditor)

import java.io.*;
import java.util.*;

public class Main {
    private static PrintWriter pw = new PrintWriter(System.out);
    private static InputReader sc = new InputReader();
    private static final int intmax = Integer.MAX_VALUE, intmin = Integer.MIN_VALUE;

    static class Pair{
        int first, second;
        Pair(int first, int second){
            this.first = first;
            this.second = second;
        }
    }

    static class InputReader{
        private static BufferedReader r = new BufferedReader(new InputStreamReader(System.in));
        private static StringTokenizer tk;

        private void next()throws IOException{
            if(tk == null || !tk.hasMoreTokens())   
                tk = new StringTokenizer(r.readLine());
        }

        private int nextInt()throws IOException{
            next();
            return Integer.parseInt(tk.nextToken());
        }

        private long nextLong()throws IOException{
            next();
            return Long.parseLong(tk.nextToken());
        }

        private String readString()throws IOException{
            next();
            return tk.nextToken();
        }

        private double nextDouble()throws IOException{
            next();
            return Double.parseDouble(tk.nextToken());
        }

        private int[] intArray(int n)throws IOException{
            next();
            int arr[] = new int[n];

            for(int i=0; i<n; i++)
                arr[i] = nextInt();

            return arr;
        }

        private long[] longArray(int n)throws IOException{
            next();
            long arr[] = new long[n];

            for(int i=0; i<n; i++)
                arr[i] = nextLong();

            return arr;
        }
    }

    public static void main(String args[])throws IOException{
        int t = sc.nextInt();
        while(t-->0)    solve();
        pw.flush();
        pw.close();
    }
    
    private static ArrayList<ArrayList<Pair>> graph;
	private static int n;
	private static long weight[];
	
	private static void graph_input(boolean isTree)throws IOException{
		graph = new ArrayList<>();
		n = sc.nextInt();
		weight = new long[n-1];
		int m = isTree ? n-1: sc.nextInt();
		
		for(int i=0; i<n; i++)
			graph.add(new ArrayList<>());
			
		for(int i=0; i<m; i++){
			int edge[] = sc.intArray(2);
			int u = edge[0] - 1, v = edge[1] - 1;
			graph.get(u).add(new Pair(v, i));
			graph.get(v).add(new Pair(u, i));
		}
	}
	
	private static int dfs(int node, int p){
		int sz = 1;
		
		for(Pair c: graph.get(node)){
			if(c.first != p){
				int csize = dfs(c.first, node);
				weight[c.second] = (long)(n-csize) * csize;
				sz += csize;
			}
		}
		
		return sz;
	}
	
	private static void reverse(long p[]){
		int i = 0, j = p.length-1;
		
		while(i < j){
			long temp = p[i];
			p[i] = p[j];
			p[j] = temp;
			i++;
			j--;
		}
	}

    private static void solve()throws IOException{
		graph_input(true);
		dfs(0, -1);
		
		Arrays.sort(weight);
		reverse(weight);
		
		int m = sc.nextInt();
		
		long p[] = sc.longArray(m), mod = (long)1e9 + 7;
		Arrays.sort(p);
		reverse(p);
		
		if(m <= n-1){
			long ans = 0;
			
			for(int i=0; i<n-1; i++){
				long q = (i >= m)?1:p[i];
				ans = (ans%mod + (q*weight[i])%mod)%mod;
			}
			
			pw.println(+ans);
		}
		else{
			int i;
			
			long ans = 0, temp = 1;
			
			for(i=0; i<=m-n+1; i++)
				temp = (temp * p[i])%mod;
				
			ans = (temp * weight[0])%mod;
			
			for(int j=1; j<n-1; j++, i++)
				ans = (ans%mod + (weight[j] * p[i])%mod)%mod;
			
			pw.println(+ans);
		}
    }
}