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 CodeForces;

import java.io.*;
import java.util.*;
 
public class D 
{	
	static ArrayList<Integer> g[];
	static int[] subtreeSize, par, vis;
	static long mod=(long)1e9+7;
	
	public static void main(String[] args) throws IOException
	{	
		FastReader f=new FastReader();
		StringBuffer sb=new StringBuffer();
	
		int test=f.nextInt();
		
		while(test-->0)
		{
			int n=f.nextInt();
			g=new ArrayList[n+1];
			for(int i=0;i<=n;i++)
				g[i]=new ArrayList<>();
			subtreeSize=new int[n+1];
			par=new int[n+1];
			vis=new int[n+1];
			
			for(int i=1;i<n;i++)
			{
				int x=f.nextInt();
				int y=f.nextInt();
				
				g[x].add(y);
				g[y].add(x);
			}
			
			dfs(1,0);
			
			
//			System.out.print("st size -> ");
//			for(int i=0;i<=n;i++)
//				System.out.print(subtreeSize[i]+" ");
//			System.out.println();
//			System.out.print("par -> ");
//			for(int i=0;i<=n;i++)
//				System.out.print(par[i]+" ");
//			System.out.println();
			List<Integer> st=new ArrayList<Integer>();
			
			for(int i=2;i<=n;i++)
				st.add(subtreeSize[i]);
			Collections.sort(st);
//			System.out.println(st);
			
			int k=f.nextInt();
			List<Integer> l1=new ArrayList<>();
			
			for(int i=0;i<k;i++)
				l1.add(f.nextInt());
			Collections.sort(l1);
			
			int val[]=new int[n-1];
			Arrays.fill(val,1);
			
			if(k>n-1)
			{
				for(int i=0;i<n-1;i++)
					val[i]=l1.get(i);
				for(int i=n-1;i<l1.size();i++)
				{	
//					System.out.println("val[n-2] = "+val[n-2]);
					val[n-2]=(int)mult(l1.get(i),val[n-2]);
				}
			}
			else
			{
				for(int i=0;i<k;i++)
					val[i]=l1.get(i);
				Arrays.sort(val);
			}
			// val ok
			// st size ok
			List<Long> times=new ArrayList<>();
			for(int i : st)
			{
				long t=(long)i*(long)(n-i);
				times.add(t);
			}
			Collections.sort(times);
			long ans=0;
			for(int i=st.size()-1;i>=0;i--)
			{	
				
				ans=(ans+mult(times.get(i),val[i]))%mod;
			}
			
			sb.append(ans+"\n");
		}
		
		System.out.println(sb);
	}
	
	static long mult(long a,long b)
	{
		return (a*b%mod);
	}
	
	static void dfs(int n,int p) 
	{
		vis[n] = 1;par[n]=p;
		subtreeSize[n]=1;
		for (int child : g[n]) 
		{
			if (vis[child] != 0)
				continue;
			
			dfs(child,n);
			subtreeSize[n]+=subtreeSize[child];
		}

	}
	
	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(); 
	    } 
	    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; 
	    } 
	} 
}
 /*
  * 
1
7
6 1
2 3
4 6
7 3
5 1
3 6
7
2 3 5 7 11 13 17
*/