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.*;
import java.lang.*;

public class D {
	public static int MOD = (int)(1e9+7);
	public static void main(String[] args) {
		FastReader in = new FastReader();
		PrintWriter out = new PrintWriter(System.out);
		int T = in.nextInt();
		for(int tt=0;tt<T;tt++) {
			int n=in.nextInt();
			Node[] nodes = new Node[n];
			for(int i=0;i<n;i++) nodes[i] = new Node();
			for(int i=1;i<n;i++) {
				int u = in.nextInt()-1;
				int v = in.nextInt()-1;
				nodes[u].adj.add(nodes[v]);
				nodes[v].adj.add(nodes[u]);
			}
			int nFactors = in.nextInt();
			int[] factors = in.readArray(nFactors);
			ruffleSort(factors);
			
			long[] finalFactors = new long[(n-1)];
			Arrays.fill(finalFactors,1);
			for(int i=0;i<Math.min(n-1, nFactors);i++) {
				finalFactors[i] = factors[i];
			}
			int index=n-1;
			while(index<nFactors) {
				finalFactors[n-2] = mul(finalFactors[n-2],factors[index]);
				index++;
			}
			nodes[0].dfs(null);
			long[] edges = new long[n-1];
			for(int i=1;i<n;i++) {
				long imp = (long)nodes[i].subtreeSize*(n-nodes[i].subtreeSize);
				edges[i-1] = imp;
			}
			ruffleSort(edges);
			ruffleSort(finalFactors);
			long ans = 0;
			for(int i=0;i<n-1;i++) {
//				out.println("Matching the edge "+edges[i]+" with factors "+finalFactors[i]);
				ans = (ans+ mul(edges[i],finalFactors[i]))%MOD;
			}
			out.println(ans);
		}
		out.close();
	}
	public static long mul(long a,long b) {
		return (a*b)%MOD;
	}
	static class Node{
		ArrayList<Node> adj = new ArrayList<>();
		int subtreeSize;
		public void dfs(Node parent) {
			this.subtreeSize = 1;
			for(Node node:adj) {
				if(node!=parent) {
					node.dfs(this);
					this.subtreeSize += node.subtreeSize;
				}
			}
		}
	}
	static final Random random=new Random();
	//	static void ruffleSort(Pair[] a) {
	//		int n=a.length;//shuffle, then sort 
	//		for (int i=0; i<n; i++) {
	//			int oi=random.nextInt(n);
	//			Pair temp=a[oi];
	//			a[oi]=a[i]; a[i]=temp;
	//		}
	//		Arrays.sort(a);
	//	}
	static void ruffleSort(int[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n), temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	static void ruffleSort(long[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			long temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	static void ruffleSort(char[] a) {
		int n=a.length;//shuffle, then sort 
		for (int i=0; i<n; i++) {
			int oi=random.nextInt(n);
			char temp=a[oi];
			a[oi]=a[i]; a[i]=temp;
		}
		Arrays.sort(a);
	}
	static class FastReader 
	{ 
		BufferedReader br; 
		StringTokenizer st; 

		public FastReader() 
		{ 
			br = new BufferedReader(new
					InputStreamReader(System.in)); 
		} 

		String next() 
		{ 
			while (st == null || !st.hasMoreElements()) 
			{ 
				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()); 
		} 

		double nextDouble() 
		{ 
			return Double.parseDouble(next()); 
		} 

		String nextLine() 
		{ 
			String str = ""; 
			try
			{ 
				str = br.readLine(); 
			} 
			catch (IOException e) 
			{ 
				e.printStackTrace(); 
			} 
			return str; 
		} 

		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
	}
}

//class Pair implements Comparable<Pair>{
//    int a;
//    int b;
//    public Pair(int a, int b) {
//        this.a = a;
//        this.b = b;
//    }
//    public int compareTo(Pair o) {
//        if(this.a==o.a)
//            return this.b - o.b;
//        return this.a - o.a;
//    }   
//}