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.*;
import java.math.BigInteger;

public class tr0 {
	static PrintWriter out;
	static StringBuilder sb;
	static int mod = 1000000007;
	static long inf = (long) 1e16;
	static int n, m;
	static ArrayList<Integer>[] ad;
	static long[][][] memo;
	static boolean f;
	static boolean vis[];
	static int[] sub;
	static char[] a;
	static ArrayList<Long> ar;

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		out = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			n = sc.nextInt();
			ad = new ArrayList[n];
			for (int i = 0; i < n; i++)
				ad[i] = new ArrayList<>();
			for (int i = 0; i < n - 1; i++) {
				int a = sc.nextInt() - 1;
				int b = sc.nextInt() - 1;
				ad[a].add(b);
				ad[b].add(a);
			}
			ar = new ArrayList<>();
			sub = new int[n];
			dfs(0, -1);
			int m = sc.nextInt();
			Collections.sort(ar);
			Collections.reverse(ar);
			long ans = 0;
			if (m <= n - 1) {
				PriorityQueue<Long> pq = new PriorityQueue<>((x, y) -> Long.compare(y, x));
				for (int i = 0; i < m; i++) {
					pq.add(sc.nextLong());
				}
				int id = 0;
				while (!pq.isEmpty()) {
					ans = ans + pq.poll() * ar.get(id++);
					ans = ans % mod;
				}
				while (id < ar.size())
					ans += ar.get(id++);
				ans %= mod;
			} else {
				PriorityQueue<Long> pq = new PriorityQueue<>();
				for (int i = 0; i < m; i++) {
					pq.add(sc.nextLong());
				}
				int id = 0;
				long[] ark = new long[n - 1];
				for (int i = 0; i < ark.length; i++)
					ark[i] = pq.poll();
				while (!pq.isEmpty()) {
					if (id == n)
						id = 0;
					ark[n - 2] = (ark[n - 2] * pq.poll()) % mod;
				}
				for (int i = 0; i < n - 1; i++) {
					ans = ans + ark[n-2-i] * ar.get(i);
					ans = ans % mod;
				}
			}
			out.println(ans);
		}
		out.flush();
	}

	static void dfs(int u, int p) {
		sub[u] = 1;
		for (int v : ad[u]) {
			if (v != p) {
				dfs(v, u);
				sub[u] += sub[v];
			}
		}
		if (p != -1)
			ar.add(sub[u] * 1l * (n - sub[u]));
	}

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

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

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

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