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 Solution{
	static PrintWriter out=new PrintWriter(System.out);
	public static void main (String[] args) throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String[] input=br.readLine().trim().split(" ");
		
		int numTestCases=Integer.parseInt(input[0]);
		while(numTestCases-->0){
		    input=br.readLine().trim().split(" ");
		    int n=Integer.parseInt(input[0]);
    		ArrayList<ArrayList<Integer>> graph=new ArrayList<>();
    		for(int i=0;i<=n;i++){
    		    graph.add(new ArrayList<>());
    		}
    		for(int i=0;i<n-1;i++){
    		    input=br.readLine().trim().split(" ");
    		    int u=Integer.parseInt(input[0]);
    		    int v=Integer.parseInt(input[1]);
    		    
    		    graph.get(u).add(v);
    		    graph.get(v).add(u);
    		}
    		int m=Integer.parseInt(br.readLine());
    		ArrayList<Integer> factors=new ArrayList<>();
    		input=br.readLine().trim().split(" ");
    		for(int i=0;i<m;i++){
    		    factors.add(Integer.parseInt(input[i]));
    		}
    		out.println(maxDistributionIndex(graph,factors));
		}
		
		out.flush();
		out.close();
	}	
	public static int maxDistributionIndex(ArrayList<ArrayList<Integer>> graph,ArrayList<Integer> factors)
	{
	    int m=factors.size();
	    int n=graph.size()-1;
	    int mod=1000000000+7;
	    
	    ArrayList<Long> count=new ArrayList<>();
	    dfs(graph,1,-1,count);
	    
	    
	    if(m<=n-1){
	        while(factors.size()<n-1){
	            factors.add(1);
	        }
	        long ans=0;
	        Collections.sort(count,Collections.reverseOrder());
	        Collections.sort(factors,Collections.reverseOrder());
	        for(int i=0;i<n-1;i++){
	            long temp=(count.get(i)*factors.get(i))%mod;
	            ans=(ans+temp)%mod;
	        }
	        return (int)(ans);
	    }
	    else{
	        long temp=1;
	        Collections.sort(factors,Collections.reverseOrder());
	        for(int i=0;i<=m-(n-1);i++){
	            temp=(temp*factors.get(i))%mod;
	        }
	        Collections.sort(count,Collections.reverseOrder());
	        long ans=(temp*count.get(0))%mod;
	        for(int i=1;i<n-1;i++){
	            temp=(count.get(i)*factors.get(m-(n-1)+i))%mod;
	            ans=(ans+temp)%mod;
	        }
	        return (int)(ans);
	    }
	}
	public static int dfs(ArrayList<ArrayList<Integer>> graph,int start,int parent,ArrayList<Long> count)
	{
	    int n=graph.size()-1;
	    int ans=1;
	    for(int i=0;i<graph.get(start).size();i++){
	        if(graph.get(start).get(i)!=parent){
	            long smallAns=dfs(graph,graph.get(start).get(i),start,count);
	            long freq=smallAns*(n-smallAns);
	            count.add(freq);
	            ans+=smallAns;
	        }
	    }
	    return ans;
	}
}