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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class Smkmp {
	
	static long mod = (long)1e9+7;
	
	static long mul(long a, long b){
		return a*b%mod;
	}
	static class Node{
		ArrayList<Node> adj = new ArrayList<>();
		int subtreeSize;
		
		public void dfs(Node node){
			
			this.subtreeSize = 1;
			
			for(Node child:adj){
				
				if(child!=node){
					
					child.dfs(this);
					this.subtreeSize+=child.subtreeSize;
				}
			}
		}
	}
	public static void main(String[] args) {
		Scanner scn = new Scanner(System.in);
		
		int t = scn.nextInt();
		while(t-->0){
			
			int n = scn.nextInt();
			
			Node[] nodes = new Node[n];
			
			for(int i=0;i<n;i++) nodes[i] = new Node();
			
			for(int i=1;i<n;i++){
				int a = scn.nextInt()-1,b = scn.nextInt()-1;
				
				nodes[a].adj.add(nodes[b]);
				nodes[b].adj.add(nodes[a]);
				
			}
			
			int nPrimes = scn.nextInt();
			
			int[] factors = new int[nPrimes];
			
			for(int i=0;i<nPrimes;i++){
				factors[i] = scn.nextInt();
			}
			
			Arrays.sort(factors);
			
			
			n--;
			long[] values = new long[n];
			Arrays.fill(values, 1);
			
			
			if(nPrimes<=n){
				
				for(int i = nPrimes-1;i>=0;i--){
					
					values[nPrimes-i-1] = factors[i];
				}
			}
			else{
				for(int i = nPrimes-1;i>=0;i--){
					
					int idx = Math.max(0, n-i-1);
					values[idx] = mul(values[idx],factors[i]);
				}
			}
			
			long[] edgeWeights = new long[n];
			
			nodes[0].dfs(null);
			
			for(int i=1;i<=n;i++){
				long times = nodes[i].subtreeSize*(long)(n+1 - nodes[i].subtreeSize);
				edgeWeights[i-1] = times;
			}
			
			Arrays.sort(edgeWeights);
			
			long ans=0;
			
			for(int i=0;i<n;i++){
				ans = (ans+mul(edgeWeights[i], values[n-1-i]))%mod;
			}
			
			System.out.println(ans%mod);
		}
		
	}
	
	public static String sortString(String inputString) 
    { 
        // convert input string to char array 
        char tempArray[] = inputString.toCharArray(); 
          
        // sort tempArray 
        Arrays.sort(tempArray); 
          
        // return new sorted string 
        return new String(tempArray); 
    }

}