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.*;
import java.io.*;

public class D_1401 {
	
	static int n;
	static ArrayList<Integer>[] adjList;
	static TreeSet<Edge> set;
	static int[] count;
	static boolean[] vis;
	
	public static int dfs(int u) {
		vis[u] = true;
		int sum = 1;
		for(int v : adjList[u])
			if(!vis[v])
				sum += dfs(v);
		return count[u] = sum;
	}
	
	public static void dfs2(int u) {
		vis[u] = true;
		for(int v : adjList[u])
			if(!vis[v]) {
				set.add(new Edge(u, v, 1l * count[v] * (n - count[v])));
				dfs2(v);
			}
	}
	
	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int t = sc.nextInt();
		while(t-->0) {
			n = sc.nextInt();
			adjList = new ArrayList[n];
			for(int i = 0; i < n; i++)
				adjList[i] = new ArrayList<>();
			for(int i = 0; i < n - 1; i++) {
				int x = sc.nextInt() - 1, y = sc.nextInt() - 1;
				adjList[x].add(y);
				adjList[y].add(x);
			}
			
			count = new int[n];
			vis = new boolean[n];
			dfs(0);
			set = new TreeSet<>();
			vis = new boolean[n];
			dfs2(0);
			
			int m = sc.nextInt();
			PriorityQueue<Integer> qu1 = new PriorityQueue<>(Collections.reverseOrder());
			for(int i = 0; i < m; i++)
				qu1.add(sc.nextInt());
			
			int MOD = (int)1e9+7;
			while(qu1.size() < n - 1)
				qu1.add(1);
			boolean flag = false;
			if(qu1.size() > n - 1)
				flag = true;
			if(flag) {
				int first = qu1.poll();
				while(qu1.size() + 1 > n - 1) {
					first = (int)((1l * first * qu1.poll()) % MOD);
				}
				
				PriorityQueue<Integer> qu = new PriorityQueue<>();
				while(!qu1.isEmpty())
					qu.add(qu1.poll());
				
				int sum = 0;
				while(!set.isEmpty()) {
					Edge e = set.pollFirst();
					if(qu.isEmpty()) {
						sum = (int)((1l * sum + (1l * (1l * e.cnt % MOD) * first) % MOD) % MOD);
					} else {
						sum = (int)((1l * sum + (1l * (1l * e.cnt % MOD) * qu.poll()) % MOD) % MOD);
					}
				}
				pw.println(sum);
			} else {
				PriorityQueue<Integer> qu = new PriorityQueue<>();
				while(!qu1.isEmpty())
					qu.add(qu1.poll());
				
				int sum = 0;
				while(!set.isEmpty()) {
					Edge e = set.pollFirst();
					sum = (int)((1l * sum + (1l * (1l * e.cnt % MOD) * qu.poll()) % MOD) % MOD);
				}
				pw.println(sum);
			}
		}
		
		pw.flush();
	}
	
	public static class Edge implements Comparable<Edge> {
		int u, v;
		long cnt;
		
		public Edge(int u, int v, long cnt) {
			this.u = u;
			this.v = v;
			this.cnt = cnt;
		}
		
		public int compareTo(Edge e) {
			if(Long.compare(cnt, e.cnt) == 0) {
				if(u == e.u)
					return v - e.v;
				return u - e.u;
			}
			return Long.compare(cnt, e.cnt);
		}
	}

	public static class Scanner {
		StringTokenizer st;
		BufferedReader br;

		public Scanner(InputStream system) {
			br = new BufferedReader(new InputStreamReader(system));
		}

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

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

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

		public int nextInt() throws IOException {
			return Integer.parseInt(next());
		}

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

		public char nextChar() throws IOException {
			return next().charAt(0);
		}

		public long nextLong() throws IOException {
			return Long.parseLong(next());
		}

		public int[] nextIntArray(int n) throws IOException {
			int[] array = new int[n];
			for (int i = 0; i < n; i++)
				array[i] = nextInt();
			return array;
		}

		public Integer[] nextIntegerArray(int n) throws IOException {
			Integer[] array = new Integer[n];
			for (int i = 0; i < n; i++)
				array[i] = new Integer(nextInt());
			return array;
		}

		public long[] nextLongArray(int n) throws IOException {
			long[] array = new long[n];
			for (int i = 0; i < n; i++)
				array[i] = nextLong();
			return array;
		}

		public double[] nextDoubleArray(int n) throws IOException {
			double[] array = new double[n];
			for (int i = 0; i < n; i++)
				array[i] = nextDouble();
			return array;
		}

		public static int[] shuffle(int[] a) {
			int n = a.length;
			Random rand = new Random();
			for (int i = 0; i < n; i++) {
				int tmpIdx = rand.nextInt(n);
				int tmp = a[i];
				a[i] = a[tmpIdx];
				a[tmpIdx] = tmp;
			}
			return a;
		}

		public boolean ready() throws IOException {
			return br.ready();
		}

		public void waitForInput() throws InterruptedException {
			Thread.sleep(3000);
		}
	}
}
