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 datatypes;

import java.io.*;
import java.util.*;

public class Main{
    public static void main(String[] args) {
        FastReader fs=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int t=1;
        t=fs.nextInt();
        for(int i=1;i<=t;i++) {
        	Main m=new Main();
            m.solve(i,out,fs);
        }
        out.close();
    }
    ArrayList<Long> PP;
    ArrayList<ArrayList<Integer>> adj;
    boolean vis[];
    int n;
    long dfs(int u,int p) {
    	long count=1;
    	for(int v:adj.get(u)) {
    		if(v!=p) {
    			long subCount=dfs(v,u);
    			PP.add( (subCount*(n-subCount)) );
    			count+=subCount;
    		}
    	}
    	return count;
    }
    public void solve(int t,PrintWriter out,FastReader fs){
    	n=fs.nextInt();
    	vis=new boolean[n+1];
    	PP=new ArrayList<Long>();
    	adj=new ArrayList<ArrayList<Integer>>();
    	for(int i=0;i<=n;i++) {
    		adj.add(new ArrayList<Integer>());
    	}
    	for(int i=1;i<n;i++) {
    		int u=fs.nextInt(),v=fs.nextInt();
    		adj.get(u).add(v);
    		adj.get(v).add(u);
    	}
    	dfs(1,-1);
    	int pf=fs.nextInt();
    	Collections.sort(PP,Collections.reverseOrder());
    	long ans=0;
    	long mod=1000000007;
    	Long p[]=new Long[pf];
    	for(int i=0;i<pf;i++) {
    		p[i]=fs.nextLong();
    	}
    	Arrays.sort(p,Collections.reverseOrder());
//    	for(int pri:p) {
//    		System.out.print(pri+" ");
//    	}
//    	System.out.println();
//    	for(long pri:PP) {
//    		System.out.print(pri+" ");
//    	}
//    	System.out.println();
    	int i=0;
    	if(pf<=n-1) {
	    	for(i=0;i<pf;i++) {
	    		ans=(mod+(ans+(PP.get(i)*p[i])%mod)%mod)%mod;
	    	}
	    	for(;i<n-1;i++) {
	    		ans=(mod+(ans+PP.get(i))%mod)%mod;
	    	}
    	}else {
    		int pi=0,in=0;
    		while(in<n-1) {
    			while(pf-pi!=(n-1)-in) {
    				pi++;
    				p[pi]=((p[pi]*p[pi-1])%mod+mod)%mod;
    			}
    			ans=(mod+(ans+(PP.get(in)*p[pi])%mod)%mod)%mod;
    			in++;
    			pi++;
    		}
    	}
    	ans=(mod+ans)%mod; 
    	out.println(ans%mod);
    }
    
} 
class FastReader{
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st=new StringTokenizer("");  
    String next(){
        while(!st.hasMoreTokens()){
            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());
    }
    int[] nextIntArray(int n){
        int a[]=new int[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    } 
    long[] nextLongArray(int n){
        long a[]=new long[n];
        for(int i=0;i<n;i++)a[i]=nextInt();
        return a;
    }
    char[] nextCharArray(){
        char a[]=next().toCharArray();
        return a;
    }
}