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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Task4 {
    ArrayList<Integer>[] adj;
    long[] subtreeSize;
    long[] edgeTraversals;
    int n;
    int mod;

    public static void main(String[] args) throws IOException {
        new Task4().solve();
    }

    private void solve() throws IOException {
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);

        mod = 1000000007;
        int t = Integer.parseInt(f.readLine());
        for (int t1 = 0; t1 < t; t1++) {
            n = Integer.parseInt(f.readLine());
            adj = new ArrayList[n];
            for (int i = 0; i < n; i++) adj[i] = new ArrayList<Integer>();
            for (int i = 0; i  < n - 1; i++) {
                StringTokenizer tokenizer = new StringTokenizer(f.readLine());
                int a = Integer.parseInt(tokenizer.nextToken()) - 1;
                int b = Integer.parseInt(tokenizer.nextToken()) - 1;
                adj[a].add(b);
                adj[b].add(a);
            }

            subtreeSize = new long[n];
            edgeTraversals = new long[n];
            buildTree(0, -1);
            //System.out.println(Arrays.toString(edgeTraversals));

            int m = Integer.parseInt(f.readLine());
            Integer[] factors = new Integer[m];
            StringTokenizer tokenizer = new StringTokenizer(f.readLine());
            for (int i = 0; i < m; i++) factors[i] = Integer.parseInt(tokenizer.nextToken());
            Arrays.sort(factors, Collections.reverseOrder());

            long multiply = 1;
            int toConnect = m - (n - 1);
            ArrayList<Integer> nums = new ArrayList<Integer>();
            for (int i = 0; i <= toConnect; i++) {
                multiply *= factors[i];
                multiply %= mod;
            }
            if (multiply != 1) nums.add((int) multiply);
            for (int i = Math.max(toConnect + 1, 0);  i < m; i++) nums.add(factors[i]);
            while (nums.size() < n - 1) nums.add(1);
            //System.out.println(nums);
            long res = 0;
            PriorityQueue<Long> edges = new PriorityQueue<Long>(Collections.reverseOrder());
            for (int i = 1; i  < n; i++) edges.add(edgeTraversals[i]);
            for (int num : nums) {
                long edge = edges.remove();
                res += (long) num * edge;
                //System.out.println(num +  " " +edge);
                res %= mod;
            }

            out.println(res);
        }

        out.close();
    }

    private void buildTree(int node, int parent) {
        subtreeSize[node] = 1;
        for (int adjacent : adj[node]) {
            if (adjacent == parent) continue;
            buildTree(adjacent, node);
            subtreeSize[node] += subtreeSize[adjacent];
        }

        if (node != 0) {
            long traversals = 0;
            long otherNodes = n - 1;
            for (int adjacent : adj[node]) {
                if (adjacent == parent) continue;
                otherNodes -= subtreeSize[adjacent];
                //traversals += (long) subtreeSize[adjacent] * (n - subtreeSize[adjacent] - 1);
                //System.out.println(node + 1 + " " + subtreeSize[adjacent] * (n - subtreeSize[adjacent] - 1));
            }
            traversals += otherNodes * ( n - otherNodes);
            edgeTraversals[node] = traversals;
        }
    }
}
