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.io.*;
public class EdD {
	static int[] subtree;
	static int[] parent;
	static Map<Integer, ArrayList<Integer>> mp;
	public static void main(String[] args) throws Exception{
		long num = 1000000007;

		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int t = Integer.parseInt(bf.readLine());
 		for(int i = 0;i<t;i++){
	 		int n = Integer.parseInt(bf.readLine());
	 		mp = new HashMap<Integer, ArrayList<Integer>>();
	 		for(int j= 1;j<=n;j++){
	 			mp.put(j,  new ArrayList<Integer>());
	 		}
	 		for(int j =0;j<n-1;j++){
	 			StringTokenizer st = new StringTokenizer(bf.readLine());
	 			int v1 = Integer.parseInt(st.nextToken());
	 			int v2 = Integer.parseInt(st.nextToken());
	 			mp.get(v1).add(v2);
	 			mp.get(v2).add(v1);
	 			
	 		}
	 		parent = new int[n+1];
	 		subtree = new int[n+1];
	 		int m = Integer.parseInt(bf.readLine());
	 		ArrayList<Integer> p = new ArrayList<Integer>();
	 		StringTokenizer st = new StringTokenizer(bf.readLine());
	 		for(int j =0;j<m;j++){
	 			p.add(Integer.parseInt(st.nextToken()));
	 		}
	 		Collections.sort(p);
	 		ArrayList<Long> coefficients = new ArrayList<Long>();
	 		if (m <= n-1){
	 			for(int j = 0;j<n-1-m;j++){
	 				coefficients.add(1L);
	 			}
	 			for(int j : p){
	 				coefficients.add((long)(j));
	 			}
	 		}
	 		else{
	 			for(int j =0;j<n-2;j++){
	 				coefficients.add((long)(p.get(j)));
	 			}
	 			long prod = 1;
	 			for(int j = n-2;j<p.size();j++){
	 				prod*=(long)(p.get(j));
	 				prod%=num;
	 			}
	 			coefficients.add(prod);
	 			
	 		}
	 		parent[1] = -1;
	 		dfs(1, -1);
	 		ArrayList<Long> subt = new ArrayList<Long>();
	 		for(int j = 2;j<=n;j++){
	 			subt.add((long)(subtree[j])*(long)((n-subtree[j])));
	 		}
	 		Collections.sort(subt);
	 		long ans = 0;
	 		for(int j = 0;j<n-1;j++){
	 			ans+= ((subt.get(j)*coefficients.get(j)) %num);
	 			ans%=num;
	 		}
	 		out.println(ans);
	 		
	 		
	 		
	 		
 		}
 		
 		
 		out.println();
	 		
 		out.close();
 		
 		
 		
 	}
	public static void dfs(int v, int p){
		for(int child : mp.get(v)){
			if (child == p)
				continue;
			parent[child] = v;
			dfs(child, v);
			subtree[v]+=subtree[child];
		}
		subtree[v]+=1;
	}
	
}
 	
 
//StringJoiner sj = new StringJoiner(" "); 
//sj.add(strings)
//sj.toString() gives string of those stuff w spaces or whatever that sequence is

 		
 		
 		
 		
	


