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.
// Working program with FastReader 
import java.io.BufferedReader; 
import java.io.IOException; 
import java.io.InputStreamReader; 
import java.util.Scanner; 
import java.util.StringTokenizer; 
import java.util.Collections; 
import java.util.ArrayList; 

public class Main 
{ 
    static ArrayList<Long> edges;
    static int n;
    static long mod =(long)(1e9+7);
    public static void main(String[] args) 
	{ 
		FastReader sc=new FastReader(); 
		StringBuffer ans=new StringBuffer();
		int test=sc.nextInt();
		while(test-->0)
		{
		    n=sc.nextInt();
		    edges=new ArrayList<>();
		    ArrayList<Integer> arr[]=new ArrayList[n+1];
		    for(int i=0;i<=n;i++)   arr[i]=new ArrayList<>();
		    for(int i=0;i<n-1;i++){
		        int a=sc.nextInt();
		        int b=sc.nextInt();
		        arr[a].add(b);
		        arr[b].add(a);
		    }
		    int fact_num = sc.nextInt();
		    ArrayList<Long> fact=new ArrayList<>(fact_num);
		    for(int i=0;i<fact_num;i++)     fact.add(sc.nextLong());
		    while(fact_num<n-1){
		        fact.add(1l);
		        ++fact_num;
		    }
		    Collections.sort(fact);
		    if(fact_num>n-1){
		      //  System.out.println("Debug1");
		        long temp=1l;
		        for(int i=n-2;i<fact.size();i++)
		        temp=temp%mod*fact.get(i)%mod;
		        fact.set(n-2,temp);
		    }
		    
		  //  System.out.println(fact);
		    dfs(1,1,arr);
		    
		    long res=0l;
		    Collections.sort(edges);
		    int i=0;
		    for(long x:edges){
		        res=(res%mod+(x%mod*fact.get(i++)%mod))%mod;
		    }
		    ans.append(res+"\n");
		}
		
	    System.out.print(ans);
	} 
	
	static long dfs(int node,int par,ArrayList<Integer>[] arr){
	    
	    long newEdge = 0l;//storing number of nodes on each subtree one by one
	    long total = 1l;//storing sum of all newEdges of a node + node itself(1)
	    for(int i=0;i<arr[node].size();i++){
	        int child = arr[node].get(i);
	        if(child != par){
	            newEdge = dfs(child,node,arr);
	            total+=newEdge;
	            edges.add(newEdge*(n-newEdge));//adding number of times an edge will be called
	        }
	    }
	    return total;
	}
	
	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; 
		} 
	} 


} 
