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 Solution {
     private static final int mod = 1000000007;
     
    public static void main(String[] args) throws Exception
    { 
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter out = new PrintWriter(System.out);
        
        int t = Integer.parseInt(br.readLine());
        
        while(t> 0){
         
          int n = Integer.parseInt(br.readLine());
          
          int m  = n-1;
	      int[] from = new int[m+1], to = new int[m+1];

		for(int i = 0; i < m; i++) { 
		    StringTokenizer st = new StringTokenizer(br.readLine());
			from[i] = Integer.parseInt(st.nextToken());
			to[i] = Integer.parseInt(st.nextToken());
		  }
		  
		  int[][] g = construct(n+1, from, to);

		  long[]  vertexCount = new long[n+1];
		  DFS(g,1,-1,vertexCount);
		  
		  int k = Integer.parseInt(br.readLine());
		  
		  int[] primeFactors = new int[k];
		  StringTokenizer st = new StringTokenizer(br.readLine());
		  
		  for(int i=0;i<k;i++)primeFactors[i] = Integer.parseInt(st.nextToken());
		  Arrays.sort(primeFactors);
		  
		  List<Long> edges = new ArrayList(); 
		  findAns(g,1,-1,0,vertexCount,edges);
		  Collections.sort(edges,Collections.reverseOrder());

		  long ans = 0;
             int j = 0, i = primeFactors.length-1;
             
		  if(edges.size() < k){
		       ans = edges.get(0) ;
		      for(j = k-1 ;j >=edges.size()-1 ;j--,i--){
		           ans = (ans*primeFactors[j])%mod;
		        
		      }
		      j=1;
		  }
             
		  for(; i>=0 && j < edges.size() ;i--){
		       
		     ans = (ans+ (edges.get(j++)*primeFactors[i])%mod)%mod;  
		  }
		  
		  while(j<edges.size()) ans = (ans+edges.get(j++))%mod;
		  out.println(ans);
		  
		  t--;
        }
        out.close();
        
    } 
    
    private static void findAns(int[][] g, int u, int p, long sumTillNow, long[] vertexCount,List<Long> edges){
         for(int v: g[u]){
              if(v  == p) continue;
              long weight = ((sumTillNow+vertexCount[u]-vertexCount[v])*vertexCount[v]);
              edges.add(weight);
              findAns(g,v,u,(sumTillNow+vertexCount[u]-vertexCount[v])%mod,vertexCount,edges);
         }
    }
    
    private static long DFS(int[][] g, int u,int p,long[] vertexCount){
        long childCount= 0;
         
         for(int v: g[u]){
              if(v  == p) continue;
              childCount+= DFS(g,v,u,vertexCount);
         }
         vertexCount[u] = childCount+1;
         return vertexCount[u];
         
    }
    
    private static int[][] construct(int n, int[] from, int[] to) {  // Graph construction
		int[][] g = new int[n][];
		int[] p = new int[n];
		for (int f : from)
			p[f]++;
		for (int t : to)
			p[t]++;
		for (int i = 0; i < n; i++)
			g[i] = new int[p[i]];
		for (int i = 0; i < from.length; i++) {
			g[from[i]][--p[from[i]]] = to[i];
			g[to[i]][--p[to[i]]] = from[i];
		}
		return g;
	}

}