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.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class NewProblem {
    public static ArrayList<Integer>[] graph;
    private static long n;
    public static ArrayList<Long> res;
    private static int dfs(int item, int prev) {
        int sum = 1;
        for (int elem:graph[item]) {
            if (elem==prev) continue;
            long result = dfs(elem, item);
            res.add(result*(n-result));
            sum+=result;
        }
        return sum;
    }
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int t = Integer.parseInt(reader.readLine());
        StringBuilder answer = new StringBuilder();
        while (t-->0) {
            n = Integer.parseInt(reader.readLine());
            StringTokenizer sToken;
            graph = new ArrayList[(int) n];
            for (int i=0; i<n; i++) graph[i] = new ArrayList<Integer>();
            for (int i=1; i<n; i++) {
                sToken = new StringTokenizer(reader.readLine());
                int x = Integer.parseInt(sToken.nextToken())-1;
                int y = Integer.parseInt(sToken.nextToken())-1;
                graph[x].add(y);
                graph[y].add(x);
            }
            int start = 0;
            for (int i=0; i<n; i++) {
                if (graph[i].size()==1) {
                    start = i; break;
                }
            }
            res = new ArrayList<>();
            dfs(start, -1);
            Collections.sort(res);
            int m = Integer.parseInt(reader.readLine());
            ArrayList<Integer> list = new ArrayList<>();
            sToken = new StringTokenizer(reader.readLine());
            for (int i=0; i<m; i++) {
                list.add(Integer.parseInt(sToken.nextToken()));
            }
            for (int i=m; i<n-1; i++) list.add(1);
            Collections.sort(list);
            long ans = 0;
            for (int i=0; i<n-2; i++) {
               long x = (res.get(i) * list.get(i))%1000000007;
               ans = (ans+x)%1000000007;
            }
            int nn = (int) (n-2);
            long x = (res.get(nn) * list.get(nn))%1000000007;
            if (m>n-2) {
                for (int i=nn+1; i<m; i++) {
                    x = (x * list.get(i))%1000000007;
                }
            }
            ans = (ans+x)%1000000007;
            answer.append(ans).append("\n");
        }
        PrintWriter pw = new PrintWriter(System.out);
        pw.print(answer);
        pw.close();
    }
}