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.math.*;
public class Solution {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		final long MOD = 1000000007L;
		int t = in.nextInt();
		for(int tt = 0; tt < t; tt++) {
			long ans = 0L;
			int n = in.nextInt();
			Node[] nodes = new Node[n];
			List<Long> s = new ArrayList<Long>();
			List<Long> c = new ArrayList<Long>();
			for(int i = 0; i < n; i++) nodes[i] = new Node(i);
			for(int i = 0; i < n - 1; i++) {
				int a = in.nextInt() - 1;
				int b = in.nextInt() - 1;
				nodes[a].list.add(b);
				nodes[b].list.add(a);
			}
			int m = in.nextInt();
			for(int i = 0; i < m; i++) {
				c.add(in.nextLong());
			}
			dfs(nodes, 0, 0, s);
			Collections.sort(s);
			Collections.sort(c);
			for(int i = 0; i < m; i++) {
				if(i >= n - 1) {
					long cur = ((c.get(n - 2) % MOD) * (c.get(i) % MOD)) % MOD;
					c.set(n - 2, cur);
				}
			}
			int cnt = Math.min(s.size(), c.size()) - 1;
			for(int i = s.size() - 1; i >= 0; i--) {
				long cur = (cnt >= 0) ? c.get(cnt) : 1L;
				cnt -= 1;
				long add = ((s.get(i) % MOD) * (cur % MOD)) % MOD;
				ans = ((ans % MOD) + (add % MOD)) % MOD;
			}
			System.out.println(ans);
		}
	}
	public static long dfs(Node[] nodes, int root, int parent, List<Long> s) {
		long sum = 1L;
		for(int i = 0; i < nodes[root].list.size(); i++) {
			int nxt = nodes[root].list.get(i);
			if(nxt == parent) continue;
			long temp = dfs(nodes, nxt, root, s);
			s.add(temp * (nodes.length - temp));
			sum += temp;
		}
		return sum;
	} 
}
class Node {
	int id;
	List<Integer> list = new ArrayList<Integer>();
	public Node(int id) {
		this.id = id;
	}
}