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.
// package first;

import java.io.*;
import java.util.*;
 
public class hello {
	static Scanner sc=new Scanner(System.in);
	public static void main (String[] args) {
		int t=1;
		t=sc.nextInt();
		while(t-->0){
		    new hello().solve();
		}
	}
	void solve(){
		int n=sc.nextInt(),i,x,y,j;
		long an=0,mod=(long)1e9+7,z;
		ArrayList<Integer> v[]=new ArrayList[n+1];
		for(i=1;i<=n;i++)	v[i]=new ArrayList();
		int c[]=new int[n+1];
		ArrayList<Long> b=new ArrayList();
		boolean vis[]=new boolean[n+1];
		Arrays.fill(vis,false);
		
		for(i=0;i<n-1;i++){
			x=sc.nextInt();
			y=sc.nextInt();
			v[x].add(y);
			v[y].add(x);
		}
		int m=sc.nextInt();
		ArrayList<Integer> val=new ArrayList();
		for(i=0;i<m;i++)	val.add(sc.nextInt());
		
		dfs(1,n,v,c,vis,b);
		while(val.size()<n-1)	 val.add(1);
		Collections.sort(val);
		Collections.sort(b);
		for(i=0;i<n-2;i++){
			z=b.get(i);
			z=(z*val.get(i))%mod;
			an=(an+z)%mod;
		}
		z=1;
		for(i=n-2;i<val.size();i++)
			z=(z*val.get(i))%mod;
		z=(z*b.get(n-2))%mod;
		an=(an+z)%mod;
		System.out.println(an);
	}
	void dfs(int s,int n,ArrayList<Integer> v[],int c[],boolean vis[],ArrayList<Long> b){
		vis[s]=true;
		c[s]=1;
		for(int i:v[s]){
			if(!vis[i]){
				dfs(i,n,v,c,vis,b);
				c[s]+=c[i];
				b.add((long)c[i]*(n-c[i]));
			}
		}
	}
}   