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.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main{

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner scn=new Scanner(System.in);
		int t=scn.nextInt();
		long mod =(long)(Math.pow(10,9)+7);
		
		while(t-->0){
			int n=scn.nextInt();
			ArrayList<ArrayList<Integer>> al =new ArrayList<>();
			for(int i=0;i<n;i++){
				ArrayList<Integer> al1=new ArrayList<>();
				al.add(al1);
			}
			int [][]arr =new int[n-1][2];
			for(int i=1;i<n;i++){
				
				int val1 =scn.nextInt();
	            int val2 =scn.nextInt();
	            val1--;
	            val2--;
	            arr[i-1][0]=val1;
	            arr[i-1][1]=val2;
	            al.get(val2).add(val1);
	            al.get(val1).add(val2);
			}
			int m=scn.nextInt();
			long []p =new long[m];
			for(int i=0;i<m;i++){
				p[i]=scn.nextLong();
				
			}
			Arrays.sort(p);
			PriorityQueue<pair>pq =new PriorityQueue<>();
			int []arr1 =new int[n];
			
//			for(int i=0;i<arr1.length;i++){
//				System.out.print(arr1[i]+" ");
//			}
//			System.out.println();
			long ans=0;
		    PriorityQueue<Long> pq1=new PriorityQueue<>(Collections.reverseOrder());
		    
		     int cnt=0;
		     int j=p.length-1;
		     
		     HashSet<Integer> hs =new HashSet<>();
		     long []arr2=new long[n];
		      dfs(0,al,arr2,hs);
//		      for(int i=0;i<arr2.length;i++){
//		    	  System.out.println(arr2[i]);
//		      }
		      long []arr3 =new long[n-1];
		      for(int i=0;i<n-1;i++){
						//System.out.println(arr[i][0]+" "+arr[i][1]+" ");
						long val =Math.min(arr2[arr[i][0]], arr2[arr[i][1]]);
						//System.out.println(val+"DA");
						long val1=(long)(n);
						arr3[i]=(val1-val)*val;
						
					}
//		      for(int i=0;i<n-1;i++){
//	    	  System.out.println(arr3[i]);
//	      }
		      Arrays.sort(arr3);
              
		       j=p.length-1;
		       while (j> n - 2){
		            p[j - 1] = (1l * p[j - 1] * p[j]) % mod;
		       j--;
		       }
		      for(int i=n-2;i>=0;i--){
		    	  long val1 =arr3[i];
		    	  long  val2=1l;
		    	  if(j>=0){
		    		   val2=p[j];
		    	  }
		    	//  System.out.print(val2+"DAd"+val1+"da");
		    	  ans+= (val2*val1);
		    	//  System.out.println(ans);
		    	  ans%=mod;
		    	  
		    	  j--;
		      }
		   //   System.out.println(j);
		  System.out.println(ans);
			
		}

	}
	public static long dfs(int s , ArrayList<ArrayList<Integer>> al, long []arr,HashSet<Integer>hs){
		 if(hs.contains(s)){
			 return arr[s];
		 }
		hs.add(s);
		long ans=0;
		for(int val : al.get(s)){
			
			ans+=dfs(val,al,arr,hs);
			
		}
		hs.remove(s);
		ans++;
		arr[s]=ans;
		return ans;
		
	}
	//need to find number of time a edge can be used visited.....
	// the greatest time a edge visited the greater value it must have...
	// how to find the most visited edge....
	// just need to count the number of nodes on the left of edge and to its right...
	// o(n2) approach remove the edge and count the element in the one component and then 
	// other..? how to find it in o(n)....
	// can be find using the remove the least edges nodes first..... and go on...
	public static class pair implements Comparable<pair>{
		int val;
		int cnt;
		public  pair(int val ,int cnt){
			this.val=val;
			this.cnt=cnt;
		}
		public int compareTo(pair o){
			return this.cnt-o.cnt;
		}
	}
	

}