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 Main
{
    LinkedList<Integer> tree[];
    int noc[];//no. of children
    int par[];
    LinkedList<Integer> q;//track back
    boolean vis[];
	public static void main(String[] args)throws Exception
	{
	    Main ob=new Main();
	    ob.fun();
	}
	public void makeTree(int ui,int vi)
	{
	    tree[ui].add(vi);
	    tree[vi].add(ui);
	}
	public void bfs()
	{
	    LinkedList<Integer> queue=new LinkedList<Integer>();
	    queue.add(1);
	    vis[1]=true;
	    
	    while(queue.size()>0)
	    {
	        int num=(int)(queue.poll());
	        noc[num]=tree[num].size();
	        if(num!=1)
	        {
	            noc[num]-=1;
	        }
	        q.addFirst(num);
	        Iterator<Integer> it=tree[num].iterator();
	        while(it.hasNext())
	        {
	            int num2=(int)(it.next());
	            if(!vis[num2])
	            {
	                par[num2]=num;
	                vis[num2]=true;
	                queue.add(num2);
	            }
	        }
	    }
	}
	public void fun()throws Exception
	{
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	    PrintWriter pw=new PrintWriter(System.out);
	    int t=Integer.parseInt(br.readLine());
	    while(t-->0)
	    {
	        int n=Integer.parseInt(br.readLine());
	        int u[]=new int[n-1];
	        int v[]=new int[n-1];
	        q=new LinkedList<Integer>();
	        tree=new LinkedList[n+1];
	        noc=new int[n+1];
	        par=new int[n+1];
	        vis=new boolean[n+1];
	        for(int i=1;i<=n;i++)
	        {
	            tree[i]=new LinkedList<Integer>();
	        }
	        for(int i=0;i<(n-1);i++)
	        {
	            String s1[]=(br.readLine()).split(" ");
	            u[i]=Integer.parseInt(s1[0]);
	            v[i]=Integer.parseInt(s1[1]);
	            makeTree(u[i],v[i]);
	        }
	        int m=Integer.parseInt(br.readLine());
	        String s2[]=(br.readLine()).split(" ");
	        int pf[]=new int[m];
	        for(int i=0;i<m;i++)
	        {
	            pf[i]=Integer.parseInt(s2[i]);
	            
	        }
	        bfs();
	        
	        Iterator<Integer> it2=q.iterator();
	        while(it2.hasNext())
	        {
	            int num3=(int)(it2.next());
	            noc[par[num3]]+=noc[num3];
	        }
	        long edges[]=new long[n-1];
	        for(int i=0;i<n-1;i++)
	        {
	            int temp;
	            if(par[u[i]]==v[i])//v[i] is the parent 
	            {
	                //edges[i]=noc[u[i]];
	                temp=(noc[u[i]]+1);
	            }
	            else
	            {
	                //edges[i]=noc[v[i]];
	                temp=(noc[v[i]]+1);
	            }
	            
	            edges[i]=(long)temp;
	            temp=n-temp;
	            edges[i]*=(long)temp;
	            //edges[i]%=1000000007;
	        }
	        Arrays.sort(edges);
	        Arrays.sort(pf);
	        long ans=0l;
	        long pf2[]=new long[n-1];
	        int in=m-1;
	        if(m>(n-1))
	        {
	            pf2[n-2]=1l ;  
	            for(int i=m-1;i>=(n-2);i--)
	            {
	                pf2[n-2]%=1000000007l;
	                pf2[n-2]*=pf[i];
	                pf2[n-2]%=1000000007l;
	            }
	            for(int i=n-3;i>=0;i--)
	            {
	                pf2[i]=pf[i];
	            }
	        }
	        else
	        {
	            for(int i=0;i<n-1;i++)
	            {
	                pf2[i]=1l;
	            }
	            in=n-2;
	            for(int i=m-1;i>=0;i--)
	            {
	                pf2[in]=pf[i];
	                in--;
	            }
	        }
	        for(int i=n-2;i>=0;i--)
	        {
	            edges[i]%=1000000007l;
	            pf2[i]%=1000000007l;
	            ans%=1000000007l;
	            long temp=edges[i]*pf2[i];
	            temp%=1000000007l;
	            ans+=temp;
	            ans%=1000000007l;
	        }
	        pw.println(ans);
	        
	    }
	    pw.flush();
	}
}