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.io.*;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    long mod = 1000000000+7;
    List<Integer>[] tree;
    int[] deg, size;
    public static void main(String[] args) throws Exception {
        Main main = new Main();
        // int t = sc.nextInt();
        int t= sc.nextInt();
        while(t-->0){
            main.solve();
        }
        out.flush();
    }

    void solve() throws Exception {
        int n = sc.nextInt();
        tree = new List[n];
        for(int i=0; i<n;i++) tree[i] = new LinkedList<>();
        long[][] edge = new long[n-1][3]; // to record the edge for sorting
        for(int i=0; i<n-1; i++){
            int u = sc.nextInt()-1, v = sc.nextInt()-1;
            tree[u].add(v); tree[v].add(u);
            edge[i][0] = u; edge[i][1] = v;
        }
        deg = new int[n]; size = new int[n];
        dfs(0, 0, 0);
        for(int i=0; i<n-1; i++){
            int u = (int)edge[i][0], v = (int)edge[i][1];
            long s1 = size[u], s2 = n-s1;
            if(deg[u]<deg[v]){
                // u is the parent of v
                s2 = size[v]; s1 = n-s2;
            }
            edge[i][2] = s1*s2;
        }
        Arrays.sort(edge,Comparator.comparingLong(o1->-o1[2]));
        int m = sc.nextInt();
        ArrayList<Long> list = new ArrayList<>();
        for(int i=0; i<m;i++) list.add(sc.nextLong());
        while(list.size()<n-1) list.add(1L);
        Collections.sort(list,Collections.reverseOrder());
        long head = list.get(0)%mod;
        int idx = 1;
        while(list.size()-idx > n-2){
            head = head*list.get(idx)%mod;
            idx++;
        }
        long ans = head*edge[0][2]%mod;
        for(int i=1; i<n-1;i++){
            long p = list.get(idx);
            ans += p*edge[i][2]%mod;
            ans %= mod;
            idx++;
        }
        ans %= mod;
        out.println(ans);
    }
    void dfs(int root, int from, int level){
        deg[root] = level;
        size[root] = 1;
        for(int to:tree[root]){
            if(to==from) continue;
            dfs(to, root, level+1);
            size[root] += size[to];
        }
    }

    int paIn(String s){return Integer.parseInt(s);}

}