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 D665 {
	
	static long mod = (long) Math.pow(10, 9) + 7;
	// dfs fields
	static List<Integer>[] adj;
	static boolean visitedNodes [];
	static boolean visitedEdges [];
	static HashMap<int[], Integer> map; 
	static ArrayList<Long> edgeWeight; 
	static int n;
	
	public static void main(String[] args) {
		Scanner scan = new Scanner(System.in);
		int t = scan.nextInt();
		for (int i = 0; i < t; i++) {
			testCase(scan, i + 1);
		}
	}

	public static void testCase(Scanner scan, int t) {
		n = scan.nextInt();
		int start[] = new int[n-1];
		int end[] = new int[n-1];
		adj = new ArrayList[n+1];
		visitedNodes = new boolean[n+1];
		visitedEdges = new boolean[n-1];
		for (int i = 0; i < adj.length; i++) {
			adj[i] = new ArrayList<Integer>();
		}
		map = new HashMap<>();
		edgeWeight = new ArrayList<>();
		for (int i = 0; i < n-1; i++) {
			start[i] = scan.nextInt();
			end[i] = scan.nextInt();
			int endpoints1 [] = {start[i], end[i]};
			int endpoints2 [] = { end[i], start[i]};
			map.put(endpoints1, i);
			map.put(endpoints2, i);
			adj[start[i]].add(end[i]);
			adj[end[i]].add(start[i]);
		}
		int m = scan.nextInt();
		ArrayList<Integer> factors = new ArrayList<Integer>();
		for (int i = 0; i < m; i++) {
			factors.add(scan.nextInt()) ;
		}


		dfs(1);
		
		Collections.sort(edgeWeight, Collections.reverseOrder());
		Collections.sort(factors, Collections.reverseOrder());
		long index = 0L;
		if(m <= n-1){
			for (int i = 0; i < n-1; i++) {
				long coeff = edgeWeight.get(i);
				if(i < m) {
					long factor = factors.get(i);
					long increment = coeff  * factor % mod;
					index = (index + increment) % mod;
				} else {
					index = (index  + coeff) % mod;
				}
			}
			System.out.println(index);
			//System.out.println(edgeWeight.size());
		} else {
			long bigfactor = 1L;
			for (int i = 0; i < m - n + 2; i++) {
				bigfactor = bigfactor * factors.get(i) % mod;
			}
			long firstCoeff = edgeWeight.get(0);
			index = (index +  firstCoeff * bigfactor % mod) % mod;
			for (int i = m - n + 2; i < m; i++) {
				long factor = factors.get(i);
				long coeff = edgeWeight.get(i - m + n - 1);
				long increment = coeff*(factor)% mod;
				index = (index+increment) % mod;
			}
			//System.out.println(edgeWeight.size());
			System.out.println(index);
		}
	
	}
	
	// recursive dfs for edge weights
	static long dfs(int v) {
		
		visitedNodes[v] = true;
		long d = 1;
		
		for(Integer i : adj[v]) {

			if(!visitedNodes[i]) {
				d += dfs(i);
			}
		}
		
		edgeWeight.add(d*(n-d));
		
		return d;
	}
}

