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.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Scanner;

public class D {
    static int mod = 1000000007;
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int t = s.nextInt();
        for(int i=0;i<t;i++){
            int n =s.nextInt();
            //creating tree like data structure
            //n+1 because 1 based indexing
            ArrayList<Integer>[] edges = new ArrayList[n+1];
            for(int j=0;j< edges.length;j++){
                edges[j] = new ArrayList<>();
            }
            for(int j=0;j<n-1;j++){
                int u = s.nextInt();
                int v = s.nextInt();

                edges[u].add(v);
                edges[v].add(u);

            }
            int m = s.nextInt();
            ArrayList<Long> p = new ArrayList();
            for(int j=0;j<m;j++){
                long val = s.nextLong();
                p.add(val);
            }
            //storing chile node from each node assuming 1 as root node
            int[] childCountDp = new int[n+1];
            methodChildCount(1,0,childCountDp,edges);
            ArrayList<Long> edgeFrequencyCount = new ArrayList<>();
            findEachEdgeOccourance(1,0,childCountDp,edges,edgeFrequencyCount, n);

            Collections.sort(edgeFrequencyCount,Collections.reverseOrder());
            Collections.sort(p, Collections.reverseOrder());

            if(edgeFrequencyCount.size()>=p.size()){
                int sizeDiff = edgeFrequencyCount.size()-p.size();
                for(int j=0;j<sizeDiff;j++){
                    p.add(1l);
                }
            }else{
                ArrayList<Long> p_temp = new ArrayList();
                long prod = 1;
                int j;
                for(j=0;j<=p.size()-edgeFrequencyCount.size();j++){
                    prod = (prod * p.get(j))%mod;
                }
                p_temp.add(prod);
                int index=1;
                for(;j<p.size();j++){
                    p_temp.add(p.get(j));
                    index++;
                }
                p = p_temp;
            }

            long ans = 0;
            for(int j=0;j<edgeFrequencyCount.size();j++){
                    ans = (ans +(((edgeFrequencyCount.get(j))%mod)*p.get(j))%mod)%mod;
            }
            System.out.println(ans);

        }
    }

    private static void findEachEdgeOccourance(int curr, int last, int[] childCountDp, ArrayList<Integer>[] edges,ArrayList<Long> edgeFrequencyCount, int totalNodes) {
        ArrayList<Integer> edge = edges[curr];
        for(int i=0;i<edge.size();i++){
            int v= edge.get(i);

            if(v==last){
                continue;
            }
            long edgeFreq = childCountDp[v]*((long)totalNodes-childCountDp[v]);
            edgeFrequencyCount.add(edgeFreq);
            findEachEdgeOccourance(v,curr,childCountDp,edges,edgeFrequencyCount,totalNodes);
        }
    }

    public static int methodChildCount(int curr, int last, int[] childCountDp, ArrayList<Integer>[] edges){

        int sum = 0;
        ArrayList<Integer> edgeList = edges[curr];
        for(int i=0;i<edgeList.size();i++){

            int v= edgeList.get(i);
            if(v==last){
                continue;
            }
            sum += methodChildCount(v,curr,childCountDp,edges);
        }

        sum = sum + 1; //including self
        childCountDp[curr] = sum;
        return sum;
    }
}
