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.
// package Quarantine;

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;

public class MaximumDistributedTree {
    static ArrayList<Integer> tree[];
    static ArrayList<Long> compo;
    static int mod=1000000007;
    static int n;
    static boolean visited[];
    public static void main(String[] args)throws IOException {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int test=Integer.parseInt(br.readLine());
        StringTokenizer st;
        StringBuilder print=new StringBuilder();
        while(test--!=0){
            n=Integer.parseInt(br.readLine());
            compo=new ArrayList<>();
            tree=new ArrayList[n+1];
            visited=new boolean[n+1];
            for(int i=1;i<=n;i++){
                tree[i]=new ArrayList<>();
            }
            for(int i=1;i<n;i++){
                st=new StringTokenizer(br.readLine());
                int x=Integer.parseInt(st.nextToken());
                int y=Integer.parseInt(st.nextToken());
                tree[x].add(y);
                tree[y].add(x);
            }
            int m=Integer.parseInt(br.readLine());
            ArrayList<Integer> primes=new ArrayList<>();
            st=new StringTokenizer(br.readLine());
            for(int i=1;i<=m;i++){
                primes.add(Integer.parseInt(st.nextToken()));
            }
            Collections.sort(primes,Collections.reverseOrder());
            ArrayList<Long> p=new ArrayList<>();
            if(m>=n){
                long f=1;
                for(int i=0;i<m-n+2;i++){
                    f=(f*primes.get(i))%mod;
                }
                p.add(f);
                for(int i=m-n+2;i<m;i++){
                    p.add(1l*primes.get(i));
                }
            }
            else{
                for(int i=0;i<m;i++){
                    p.add(1l*primes.get(i));
                }
                for(int i=m;i<n;i++){
                    p.add(1l);
                }
            }
            dfs(1);
            Collections.sort(compo,Collections.reverseOrder());
            long ans=0;
            for(int i=0;i<n-1;i++){
                long temp=(p.get(i)*compo.get(i))%mod;
                ans=(ans+temp)%mod;
            }
            print.append(ans).append("\n");
        }
        System.out.print(print.toString());
    }
    public static int dfs(int curr){
        visited[curr]=true;
        int size=1;
        for(int j:tree[curr]){
            if(!visited[j]){
                int down=dfs(j);
                int up=n-down;
                compo.add(1l*down*up);
                size+=down;
            }
        }
        return size;
    }
}
