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 Graph
{
    static ArrayList<Integer> adj[];
    static long dp[];
    static long M=1000000007;
    static int V;
    Graph(int v)
    {
        V=v;
        adj=new ArrayList[v];
        for(int i=0;i<v;i++)
        adj[i] = new ArrayList<>();
    }
    void addEdge(int u,int v)
    {
        adj[u].add(v);
        adj[v].add(u);
    }
    
    void DFS(){
       boolean v[]=new boolean[V];
       
       DFSUtil(0,v);
        
    }
    
    long DFSUtil(int s,boolean visited[]){
        visited[s]=true;
        long cnt=0;
        for(int i=0;i<adj[s].size();i++){
            
            if(!visited[adj[s].get(i)]){
                cnt=cnt+DFSUtil(adj[s].get(i),visited);
                
            }
            
        }
        cnt=cnt+1;
        dp[s]=((cnt)*((long)V-cnt));
        return cnt;
    }
    
    public static void main (String[] args)throws java.lang.Exception {
     // Scanner sc=new Scanner(System.in);
	    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
	   // int t=1;
	    //t=sc.nextInt();
	    int t=Integer.parseInt(br.readLine());
	    while(--t>=0){
	                int n=Integer.parseInt(br.readLine());
	                dp=new long[n];
	                StringTokenizer st;
	                Graph g=new Graph(n);
	                for(int i=1;i<n;i++){
	                    st=new StringTokenizer(br.readLine());
	                    g.addEdge(Integer.parseInt(st.nextToken())-1,Integer.parseInt(st.nextToken())-1);
	                }
	                int m=Integer.parseInt(br.readLine());
	                st=new StringTokenizer(br.readLine());
	                long arr[]=new long[m];
	                for(int i=0;i<m;i++){
	                    arr[i]=Long.parseLong(st.nextToken());
	                }
	                g.DFS();
	                
	                long ans[]=new long[n];
	                Arrays.sort(arr);
	                Arrays.sort(dp);
	                
	                Arrays.fill(ans,1);
	                for(int i=n-1;i<m;i++){
	                    arr[n-2]=(arr[n-2]%M*arr[i]%M)%M;
	                }
	                int j=Math.min(m-1,n-2);
	                for(int i=n-1;i>0;i--){
	                    if(j>=0){
	                    ans[i]=(dp[i]%M*arr[j]%M)%M;
	                    j--;
	                    }
	                    else{
	                        ans[i]=dp[i];
	                    }
	                }
	                //System.out.println(dp[0]);
	                long res=0;
	                for(int i=1;i<n;i++){
	                    res=(res+ans[i])%M;
	                }
	                System.out.println(res);
	        
	    }
    }
    
}