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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Main
{
	static ArrayList<ArrayList<Integer>> g ;
	static long dp[] ,N ;
	static ArrayList<Long> val ;
	public static void main(String[] args) {
		new Thread(null, new Runnable() {
			public void run() {
                solve();
            }
        }, "1", 1 << 26).start();
	}
	static void solve () {
		FastReader fr =new FastReader();	PrintWriter op =new PrintWriter(System.out);
 		
 		g =new ArrayList<>() ;	dp =new long[(int)1e5 + 5] ;	val =new ArrayList<>() ; 
 		int t =fr.nextInt() ,n ,m ,i ,j ,k ,l ;
 		long p[] ,md =1000000007l ,lft ,rht ,ans ;
 		
 		while (t-- > 0) {
 			N =n =fr.nextInt() ;
 			for (i =0 ; i<n ; ++i)	g.add (new ArrayList<>()) ;
 			for (i =1 ; i<n ; ++i) {
 				j =fr.nextInt()-1 ;	k =fr.nextInt()-1 ;
 				g.get(j).add(k) ;	g.get(k).add(j) ;
 			}
 			
 			m =fr.nextInt() ;	p =new long[(l =Math.max(n-1 , m))] ;
 			Arrays.fill(p,1) ;	Arrays.fill(dp,0) ;
 			
 			for (i =0 ; i<m ; ++i)	p[i] =fr.nextLong() ;
 			dfs(0,-1) ;	ans =0l ;
 		//for (i =0 ; i<n ;++i)System.out.println(dp[i]) ;
 			Collections.sort (val) ;	sort (p,0,l-1) ;
 			for (i =0 ; i<n-2 ; ++i) {
 				lft =val.get(i)%md ;	rht =p[i]%md ;
 				lft =(lft*rht)%md ;
 				ans =(ans + lft)%md ;
 			}

 			lft =val.get(i)%md ;	rht =1l ;
 			for (; i<l ; ++i)	rht =(rht*p[i])%md ;
 			lft =(lft*rht)%md ;
 			ans =(ans + lft)%md ;

 			op.println(ans) ;
 			g.clear() ;	val.clear() ;
 		}
		op.flush();	op.close();
	}
	static void dfs (int n , int p) {
		int i ,j ;	dp[n] =1l ;	long dm ;

		for (i =0 ; i<g.get(n).size() ; ++i) {
			j =g.get(n).get(i) ;	if (j==p)	continue;
			
			dfs(j,n) ;	dp[n] += dp[j] ;

			dm =dp[j] * (N - dp[j]) ;	val.add(dm) ;
		}
	}
	public static void sort(long[] arr , int l , int u) {
		int m ;
 
		if(l < u){
			m =(l + u)/2 ;
 
			sort(arr , l , m);	sort(arr , m + 1 , u);
 
			merge(arr , l , m , u);
		}
	}
	public static void merge(long[] arr , int l , int m , int u) {
		long[] low = new long[m - l + 1] ;
 
		long[] upr = new long[u - m] ;
 
		int i ,j =0 ,k =0 ;
 
		for(i =l;i<=m;i++)
			low[i - l] =arr[i] ;
 
		for(i =m + 1;i<=u;i++)
			upr[i - m - 1] =arr[i] ;

		i =l;
 
		while((j < low.length) && (k < upr.length))
		{
			if(low[j] < upr[k])
				arr[i++] =low[j++] ;
			else 
				arr[i++] =upr[k++] ;
		}
 
		while(j < low.length)
			arr[i++] =low[j++] ;
 
		while(k < upr.length)
			arr[i++] =upr[k++] ;
	}
	static class FastReader {
		BufferedReader br;
		StringTokenizer st;

		public FastReader() {
			br =new BufferedReader(new InputStreamReader(System.in));
		}

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

		String nextLine() {
			String str ="";

			try
			{
				str =br.readLine();
			}
			catch(IOException e)
			{
				e.printStackTrace();
			}

			return str;
		}

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

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