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.*;
import java.util.*;

public class d
{
    public static void print(String str,long val){
        System.out.println(str+"  "+val);
    }

    public long gcd(long a, long b) {
        if (b==0L) return a;
        return gcd(b,a%b);
    }

    public static void debug(long[][] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.println(Arrays.toString(arr[i]));
        }
    }
    public static void debug(int[][] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.println(Arrays.toString(arr[i]));
        }
    }
    public static void debug(String[] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.println(arr[i]);
        }
    }

    public static void print(int[] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.print(arr[i]+" ");
        }
        System.out.print('\n');
    }

    public static void print(Object[] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.print(arr[i]+" ");
        }
        System.out.print('\n');
    }
    public static void print(String[] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.print(arr[i]+" ");
        }
        System.out.print('\n');
    }
    public static void print(long[] arr){
        int len = arr.length;
        for(int i=0;i<len;i++){
            System.out.print(arr[i]+" ");
        }
        System.out.print('\n');
    }

    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }
        public FastReader(String path) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(path));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException  e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {
            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }

    static long mod = (long)(1e9+7);
    public static void main(String[] args)
    {
        FastReader s=new FastReader();
        int t = s.nextInt();
        for(int tt=0;tt<t;tt++){
            int n = s.nextInt();
            ArrayList<Integer>[] list = new ArrayList[n];
            for(int i=0;i<n;i++){
                list[i] = new ArrayList<>();
            }
            for(int i=0;i<n-1;i++){
                int u = s.nextInt()-1;
                int v = s.nextInt()-1;
                list[u].add(v);
                list[v].add(u);
            }
            int m = s.nextInt();
            Integer[] primes = new Integer[m];
            for(int i=0;i<m;i++){
                primes[i] = s.nextInt();
            }
            System.out.println(solve(n,list,m,primes));
        }

    }

    static long solve(int n,ArrayList<Integer>[] list,int m,Integer[] primes){
        Arrays.sort(primes,Collections.reverseOrder());
        int[] subtree = new int[n];
        dfs(0,-1,list,subtree);
        ArrayList<Long> n_times = new ArrayList<>();
        count(n,0,-1,subtree,n_times,list);
        Collections.sort(n_times,Collections.reverseOrder());

//        System.out.println(n_times);
//        print(subtree);
//        print(primes);
        if(n-1>=m){
            long ans =0;
            for(int i=0;i<m;i++){
                ans+=(n_times.get(i)*primes[i]);
                ans%=mod;
            }
            for(int i=m;i<n-1;i++){
                ans+=n_times.get(i);
                ans%=mod;
            }
            return ans;
        }
        else {
            long prdt =1;
            for(int i=0;i<=(m+1-n);i++){
                prdt*=primes[i];
                prdt%=mod;
            }
            long ans = prdt*n_times.get(0);
            ans%=mod;
            for(int i=m+2-n;i<m;i++){
                ans+=(primes[i]*n_times.get(i-(m+1-n)));
                ans%=mod;
            }
            return ans;
        }



    }

    static void dfs(int curr,int parent,ArrayList<Integer>[] list,int[] subtree){
        if(list[curr].size()==1 && parent==list[curr].get(0)){
            subtree[curr] =1;
            return;
        }
        for(int child:list[curr]){
            if(child!=parent){
                dfs(child,curr,list,subtree);
            }
        }
        subtree[curr]+=1;
        for(int child:list[curr]){
            if(child!=parent){
                subtree[curr]+=subtree[child];
            }
        }
    }

    static void count(int n,int curr,int parent,int[] subtree,ArrayList<Long> n_times,ArrayList<Integer>[] list){
        for(int child:list[curr]){
            if(child!=parent){
                long temp = subtree[child]*1l*(n-subtree[child]);
                n_times.add(temp);
                count(n,child,curr,subtree,n_times,list);
            }
        }
    }




//    OutputStream out = new BufferedOutputStream( System.out );
//        for(int i=1;i<n;i++){
//              out.write((arr[i]+" ").getBytes());
//         }
//        out.flush();
//  long start_time = System.currentTimeMillis();
//  long end_time = System.currentTimeMillis();
//  System.out.println((end_time - start_time) + "ms");



}
