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.*;
public class D {
	static ArrayList<Integer>[] a;
	static boolean[] visited;
	static ArrayList<Long> nums;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		long t = in.nextInt();
		for(long T=0;T<t;T++) {
			int n = in.nextInt();
			a = new ArrayList[n];
			visited = new boolean[n];
			nums = new ArrayList<Long>();
			for(int i=0;i<n;i++) {
				a[i] = new ArrayList<Integer>();
			}
			for(int i=0;i<n-1;i++) {
				int x = in.nextInt()-1;
				int y = in.nextInt()-1;
				a[x].add(y);
				a[y].add(x);
			}
//			for(long i=0;i<n;i++) {
//				System.out.prlong(i+1+": ");
//				for(long j:a[i]) {
//					System.out.prlong(j+1+", ");
//				}
//				System.out.prlongln();
//			}
			visited[0] = true;
			Node s = assign(0,0);
			count(s, 0);
//			System.out.prlongln(nums);
			long m = in.nextInt();
			ArrayList<Long> c = new ArrayList<Long>();
			for(long i=0;i<m;i++) {
				c.add(in.nextLong());
			}
			while(c.size()<n-1) {
				c.add((long)1);
			}
			Collections.sort(nums);
			Collections.sort(c);
			while(c.size()>n-1) {
				long x = c.remove(c.size()-1);
				long y = c.remove(c.size()-1);
				c.add((x*y)%1000000007);
			}
//			System.out.prlongln(nums);
//			System.out.prlongln(c);
			long total = 0;
			for(int i=0;i<n-1;i++) {
				total = (total+c.get(c.size()-1-i)*nums.get(nums.size()-1-i))%1000000007;
			}
			System.out.println(total);
		}
		
	}
	static void count(Node n, long s) {
		long total = 0;
		for(Node i:n.children) {
			total+=i.total;
		}
		for(Node i:n.children) {
			long x = (long)i.total*(total-i.total+s+i.d);
			nums.add(x);
//			System.out.prlongln(i.i+" "+x);
			count(i, total-i.total+s);
		}
	}
	
	static Node assign(int x, long depth) {
		Node n = new Node();
		n.d = depth;
		n.i=x;
		for(int i:a[x]) {
			if(!visited[i]) {
				visited[i] = true;
				Node temp = assign(i, depth+1);
				n.total+=temp.total%1000000007;
				n.children.add(temp);
			}
		}
//		System.out.prlongln(x+" "+depth+" "+n.total);
		return n;
	}
	
	
	static class Node {
		long i;
		ArrayList<Node> children = new ArrayList<Node>();
		long total = 1;
		long d=0;
	}
}