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.awt.List;
import java.io.*;


public class Main {
	static ArrayList<Long> h;
    public static void main(String[] args) throws Exception {
 
    	Scanner s=new Scanner(System.in);
 
    	int t=s.nextInt();
    	long m=1000000007;
    	for(int ie=0;ie<t;ie++) {
    		int n=s.nextInt();
    		ArrayList[] arr=new ArrayList[n+1];
    		for(int i=0;i<n+1;i++) {
    			ArrayList<Integer> list=new ArrayList<>();
    			arr[i]=list;
    		}
    		
    		for(int i=0;i<n-1;i++) {
    			int a=s.nextInt();
    			int b=s.nextInt();
    			arr[a].add(b);
    			arr[b].add(a);
    		}
    		
    		int me=s.nextInt();
    		Integer[] p=new Integer[me];
    		for(int i=0;i<me;i++) {
    			p[i]=s.nextInt();
    		}
    		Arrays.sort(p);
    		ArrayList<Long> l=new ArrayList<>();
    		if(n-1<me) {
    			long anss=1;
    			int a=me-n+1;
    			for(int i=0;i<me-a-1;i++) {
    				l.add(p[i]*1l);
    						
    			}
    			for(int i=me-a-1;i<me;i++) {
    				anss=(anss%m*p[i]%m)%m;
    			}
    			
    			l.add(anss);
    		}else {
    			for(int i=0;i<me;i++) {
    				l.add(p[i]*1l);
    			}
    		}
    		
    		 h=new ArrayList<>();
    		 int[] visited=new int[n+1];
    		dfs(arr,visited,1,n,m);
    		
    		Collections.sort(h);
    		long ans=0;
    		
    		int st=l.size()-1;
    	//	System.out.println(l);
    		//System.out.println(h);
    		for(int i=h.size()-1;i>=0;i--) {
    			if(st<0) {
    				ans=(ans%m+h.get(i)%m)%m;
    			}else {
    				
    			ans=(ans%m+(h.get(i)%m*l.get(st)%m*1l)%m)%m;
    			}
    			
    			st--;
    		}
    		
    		System.out.println(ans);
    		
    		
    	}
    	
    	
    }
    public static long dfs(ArrayList[] arr,int[] visited,int start,int n,long m) {
    	visited[start]=1;
    	long g=1;
    	for(int i=0;i<arr[start].size();i++) {
    		int curr=(int)arr[start].get(i);
    		if(visited[curr]==0) {
    			g=g+dfs(arr,visited,curr,n,m);
    		}
    	}
    	
    	if(start!=1)
    	h.add((1l*g*Math.abs((n-g))));
    	
    	
    	return g;
    }
    
    
}