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.*; 


public class _1401D { 
   public static void main(String[] args)  { 
	   Scanner input = new Scanner(System.in);
	   long base = (long) (1e9 + 7);
	   int t = input.nextInt();
	   while (t -- > 0) {
		   int n = input.nextInt();
		   
		   Graph graph = new Graph(n);
		   for (int i = 1 ; i < n ; ++ i) {
			   int u = input.nextInt();
			   int v = input.nextInt();
			   graph.add(u,v);
		   }
		   
		   int m = input.nextInt();
		   long[] p = new long[m+5];
		   for (int i = 1 ; i <= m ; ++ i)
			   p[i] = input.nextInt();
		   
		   int root = 1;
		   for (int i = 1 ; i <= n ; ++ i)
			   if (graph.getEdSize(i)== 1) 
				   root = i;
		   graph.dfs(root,0);
		   Arrays.sort(graph.total, 1 , n + 1);
		  
		  
		   Arrays.sort(p, 1 , m + 1);
		   for(int i = n; i <= m; i ++) p[n - 1] = p[n - 1] * p[i] % base;
			
			int indexP =  m < n - 1 ? m : n - 1;
			long res = 0;
			
			for(int i = n; i >= 2; i --) {
				if(indexP == 0) res = (res + graph.total[i]) % base;
				else res = (res + graph.total[i] * p[indexP --]) % base; 
			}
		  
		   System.out.println(res);
	   }
   }
   
   static class Graph {
		private int n;
		private ArrayList<ArrayList<Integer>> ed =  new ArrayList<ArrayList<Integer>>();
		public long[] total;
		private int[] nChild;
		public Graph(int n) {
			this.n = n;
			this.ed = new ArrayList<ArrayList<Integer>>();
			this.total = new long[n+5];
			this.nChild = new int[n+5];
			for (int i = 0 ; i <= n ; ++ i)
				ed.add(new ArrayList<Integer>());
		}
		
		public void add(int u, int v) {
			ed.get(u).add(v);
			ed.get(v).add(u);
		}
		
		public int getEdSize(int u) {
			return ed.get(u).size();
		}
		
		void dfs(int u, int pr) {
			nChild[u] = 1;
			for (int v : ed.get(u)) {
				if (v == pr) continue;
				dfs(v,u);
				nChild[u] += nChild[v];
				total[v] = (long)1*nChild[v]*(n - nChild[v]);
			}
		}
		
		
	}
}