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 LeetCode;

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

public class Main {

	static class Reader {
		BufferedReader br;
		StringTokenizer st;

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

		public int nextInt() throws IOException {
			if (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return Integer.parseInt(st.nextToken());
		}

		public float nextFloat() throws IOException {
			if (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return Float.parseFloat(st.nextToken());
		}

		public long nextLong() throws IOException {
			if (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return Long.parseLong(st.nextToken());
		}

		public double nextDouble() throws IOException {
			if (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return Double.parseDouble(st.nextToken());
		}

		public String nextLine() throws IOException {
			return br.readLine();
		}

		public String next() throws IOException {
			if (st == null || !st.hasMoreTokens()) {
				st = new StringTokenizer(br.readLine());
			}
			return st.nextToken();
		}
	}

	static class Pair<T1 extends Comparable<T1> ,T2 extends Comparable<T2>> implements Comparable<Pair<T1,T2>>{
		T1 first;
		T2 second;

		public Pair(T1 first, T2 second) {
			this.first = first;
			this.second = second;
		}
		
		@Override
		public int compareTo(Pair<T1,T2> o) {
			return this.first.compareTo(o.first);
		}	
	}
	
	public static long[] initArray(int n) throws IOException {
		long arr[] = new long[n];
		for(int i=0;i<n;i++) {
			arr[i] = sc.nextLong();
		}
		return arr;
	}
	
	public static long mul(long a, long b) {
		return (a*b)%mod;
	}
	
	public static long add(long a, long b) {
		return (a+b)%mod;
	}
	
	public static long max(long arr[]) throws IOException {
		long max = Long.MIN_VALUE;
		for(long a : arr) {
			max = Math.max(max, a);
		}
		return max;
	}
	
	public static long min(long arr[]) throws IOException {
		long min = Long.MAX_VALUE;
		for(long a : arr) {
			min = Math.min(min, a);
		}
		return min;
	}
	
	public static long sum(long arr[]) throws IOException {
		long sum = 0;
		for(long a : arr) {
			sum+=a;
		}
		return sum;
	}
	
	public static ArrayList<Integer>[] initAdjacency(int n, int e, boolean undirected) throws IOException{
		ArrayList<Integer>[] list = new ArrayList[n+1];
		for(int i=0;i<=n;i++) {
			list[i] = new ArrayList<>();
		}		
		for(int i=0;i<e;i++) {
			int u = sc.nextInt();
			int v = sc.nextInt();
			list[u].add(v);
			if(undirected)
				list[v].add(u);
		}
		return list;
	}

	private static void solve() throws IOException {
		int n = sc.nextInt();
		ArrayList<Integer>[] graph = initAdjacency(n, n-1, true);
		int m = sc.nextInt();
		long[] k = initArray(m);
		Arrays.sort(k);
		
		long[] weights = new long[n-1];
		if(m>=n-1) {
			for(int i=0;i<n-1;i++)
				weights[i] = k[i];
			for(int i=n-1;i<m;i++)
				weights[n-2] = mul(weights[n-2], k[i]);
		} else {
			int index = n-2;
			for(int i=m-1;i>=0;i--) {
				weights[index--] = k[i]; 
			}
			while(index>=0) {
				weights[index--] = 1;
			}
		}
		
		values = new long[n-1];
		index = 0;
		
		dfs(graph,0,1);
		Arrays.sort(values);
		
		long ans = 0;
		for(int i=0;i<n-1;i++) {
			ans = add(ans, mul(values[i]%mod, weights[i]));
		}
		out.println(ans);
	}
	
	public static long dfs(ArrayList<Integer>[] graph, int last, int curr) {
		long ans = 1;
		for(int j : graph[curr]) {
			if(j!=last) {
				long sub = dfs(graph, curr, j);
				long temp = sub*(graph.length-1-sub);
				values[index++] = temp;
				ans+=sub;
			}
		}
		return ans;
	}
	
	static long[] values;
	static int index;
	
	private static void readt() throws IOException {
		T = sc.nextInt();
	}

	static int mod = (int) (1e9 + 7);
	static int T;
	static Reader sc;
	static PrintWriter out;
	public static void main(String[] args) throws IOException {
		sc = new Reader();
		out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		try {
			T = 1;
			readt();
			while (T-- > 0) {
				solve();
			}
			out.flush();
		} catch (Exception e) {
			e.printStackTrace();
		}
	}
}