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 D {
    public static Node[] nodes;
    public static LinkedList<Integer>[] edges;
    public static int n;
    public static long MOD = 1000000007;
    public static void main(String[] args) throws IOException{
        BufferedReader f = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
        int t = Integer.parseInt(f.readLine());
        while(t-->0){
            n = Integer.parseInt(f.readLine());
            nodes = new Node[n+1];
            for(int i = 0; i <= n; i++) nodes[i] = new Node(i);
            edges = new LinkedList[n+1];
            for(int i = 0; i < n+1; i++) edges[i] = new LinkedList<Integer>();
            for(int i = 0; i < n-1; i++){
                StringTokenizer st = new StringTokenizer(f.readLine());
                int a = Integer.parseInt(st.nextToken());
                int b = Integer.parseInt(st.nextToken());
                edges[a].add(b);
                edges[b].add(a);
            }
            boolean[] seen = new boolean[n+1];
            Stack<Integer> stk = new Stack<Integer>();
            stk.add(1);
            int currid = 0;
            while(!stk.isEmpty()){
                int temp = stk.pop();
                if(seen[temp]) continue;
                seen[temp] = true;
                for(int e: edges[temp]){
                    if(!seen[e]){
                        nodes[temp].children.add(new Edge(currid, e));
                        currid++;
                        stk.add(e);
                    }
                }
            }
            dfs(1);
            ArrayList<Long> factorization = new ArrayList<Long>();
            int k = Integer.parseInt(f.readLine());
            StringTokenizer st = new StringTokenizer(f.readLine());
            for(int i = 0; i < k; i++) factorization.add(Long.parseLong(st.nextToken()));
            ArrayList<Edge> edges = new ArrayList<Edge>();
            for(Node node: nodes) for(Edge edge: node.children) edges.add(edge);
            long ans = 0;
            if(k < n-1){
                for(int i = 0; i < n-k; i++) factorization.add(1L);
                Collections.sort(edges, Collections.reverseOrder());
                Collections.sort(factorization, Collections.reverseOrder());
                for(int i = 0; i < n-1; i++){
                    ans+=(factorization.get(i)*edges.get(i).paths);
                    ans%=MOD;
                }
                out.println(ans);
            }else{
                Collections.sort(edges);
                Collections.sort(factorization);
                for(int i = 0; i < n-2; i++){
                    ans+=(factorization.get(i)*edges.get(i).paths);
                    ans%=MOD;
                }
                long finprod = 1L;
                for(int i = n-2; i < k; i++){
                    finprod*=factorization.get(i);
                    finprod%=MOD;
                }
                ans+=(finprod*edges.get(n-2).paths);
                ans%=MOD;
                out.println(ans);
            }



        }

        out.close();
    }
    public static void dfs(int node){
        nodes[node].size = 1;
        for(Edge e: nodes[node].children){
            dfs(e.dest);
            nodes[node].size+=nodes[e.dest].size;
            e.paths = ((long)n - (long)(nodes[e.dest].size))*((long)nodes[e.dest].size);
        }
    }
    static class Node{
        public int id;
        public ArrayList<Edge> children;
        public int size;
        public Node(int id){
            this.id=id;
            this.children = new ArrayList<Edge>();
            this.size = 0;
        }

    }
    static class Edge implements Comparable<Edge>{
        public int id;
        public int dest;
        public long paths;
        public Edge(int i, int d){
            id=i;
            dest=d;
            paths=0;
        }
        public int compareTo(Edge e){
            if(paths < e.paths) return -1;
            if(paths > e.paths) return 1;
            return 0;
        }
    }
}
