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

public class A1 {
	
	public static int MAX = 100005;
	
	public static void main(String[] args) {
		Scanner input = new Scanner(System.in);
		int mod = (int)1e9 + 7;
		int test = input.nextInt(); 
		
		while (test -- > 0) {
			int n = input.nextInt();
			Graph graph = new Graph(n);
			
			for (int i = 1; i <= n - 1; i ++) {
				int u, v;
				u = input.nextInt();
				v = input.nextInt(); 
				graph.add(u , v);
			}
			
			long[] b = graph.get();
			//graph.in();
			
			//for (int i = 1; i <= n; i ++) System.out.print(b[i] + " ");
			//System.out.println();
			
			int m = input.nextInt();
			long[] p = new long[m+5]; 
			for (int i = 1; i <= m; i ++) p[i] = input.nextLong();
			Arrays.sort(p, 1, m + 1);
			
			for (int i = n; i <= m; i ++) {
				p[n-1] *= p[i];
				p[n-1] %= mod;
			}
			
 			
			int id = m < n ? m : n - 1;
			
			//for (int i = 1; i <= id; i ++) System.out.print(p[i] + " ");
			//System.out.println();
			
			long ans = 0;
			
			for (int i = n; i >= 2; i --) {
				if (id == 0) {
					ans += b[i] % mod;
					ans %= mod;
				}
				else {
					ans += b[i] % mod * p[id] % mod;
					ans %= mod;
					id--;
				}
			}
			
			System.out.println(ans);
		}
		
		input.close();
	}
	
	static class Graph{
		private int n;
		private ArrayList<Integer>[] vc;
		private int[] sl;
		
		public Graph(int n) {
			this.n = n;
			vc = new ArrayList[n + 5];
			sl = new int[n + 5];
			for (int i = 1; i <= n; i ++) {
				vc[i] = new ArrayList<>();
			}
		}
		
		public void add(int u, int v) {
			vc[u].add(v);
			vc[v].add(u);
		}
		
		private void lui(int u, int p) {
			sl[u] = 1;
			for (int v : vc[u]) {
				if (v == p) continue;
				lui(v, u);
				sl[u] += sl[v];
			}
		}
		
		public long[] get() {
			lui(1, 0);
			long[] res = new long[n + 5];
			
			for (int i = 1; i <= n; i ++) res[i] = 	(long)sl[i] * (n - sl[i]);
			Arrays.sort(res, 1, n + 1);
			return res;
		}
		
		public void in() {
			for (int i = 1; i <= n; i++) System.out.println(" - " + sl[i]);
		}
	}
}



