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.lang.reflect.Array;
import java.math.BigInteger;
import java.text.DateFormat;
import java.text.Format;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.time.LocalDateTime;
import java.util.*;
import java.io.*;
import java.math.*;
import java.security.*;
import java.text.*;
import java.util.*;
import java.util.concurrent.*;
import java.util.function.*;
import java.util.regex.*;
import java.util.stream.*;
import static java.util.stream.Collectors.joining;
import static java.util.stream.Collectors.toList;

public class file {



    private static final Scanner sc = new Scanner(System.in);
    private static String s;
    private static int[] arr;
    private static HashSet<Long> set;
    private static HashMap<Character, Integer> cmap;
    private static Map<Long, Long> map;
    private static Stack<Integer> stk;
    private static ArrayList<Integer>[] graph;
    private static boolean[] visited;
    private static int[] population;
    private static int[] hap_index;
    private static int[] happy;
    private static ArrayList<Integer> list;
    private static ArrayList<Long> Llist;

    private static final int MOD = 1000000007;

    //find max of array
    private static int findMax(int[] arr) {
        int max = Integer.MIN_VALUE;
        for (int x : arr) {
            max = Math.max(x, max);
        }
        return max;
    }

    // print an array
    private static void printArray(int[] arr) {
        for (int x : arr) {
            System.out.print(x + " ");
        }
        System.out.println();
    }

    //DFS
    private static int dfs(int curr, int prev) {
        int count = 1;
        for (int next : graph[curr]) {
            if (next==prev) continue;
            long new_count = dfs(next, curr);
            Llist.add(new_count* (graph.length - new_count));
            count += new_count;
        }
        return count;
    }

    private static void foo(int testCase) {
        int n = sc.nextInt(), a, b;
        graph = new ArrayList[n];
        for (int i = 0; i < n; i++) graph[i] = new ArrayList<>();
        for (int i = 0; i < n - 1; i++) {
            a = sc.nextInt() - 1;
            b = sc.nextInt() - 1;
            graph[a].add(b);
            graph[b].add(a);
        }
        int m = sc.nextInt();
        list = new ArrayList<>();
        for (int i = 0; i < m; i++) list.add(sc.nextInt());
        for (int i = m; i < n - 1; i++) list.add(1);
        Collections.sort(list);
        //find the root
        int root = 0;
        for (int i = 0; i < n; i++) {
            if (graph[i].size() == 1) {
                root = i;
                break;
            }
        }
        Llist = new ArrayList<>();
        dfs(root, -1);
        Collections.sort(Llist);
        long ans = 0, x;
        for (int i = 0; i < n - 2; i++) {
            x = (Llist.get(i)*list.get(i))%MOD;
            ans = (ans + x)%MOD;
        }
        x = (Llist.get(n - 2)*list.get(n-2))%MOD;
        if (m > n - 2) {
            for (int i = n - 1; i < m; i++)
                x = (x * list.get(i))%MOD;
        }
        ans = (ans + x)%MOD;
        System.out.println(ans);
    }



    public static void main(String[] args) throws IOException {
        //foo(0);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            foo(i);
        }
    }
}