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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.List;
import java.util.StringTokenizer;

public class R665D {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		int t = scn.nextInt();
		while (t > 0) {
			int n = scn.nextInt();
			List<Edge>[] tree = new ArrayList[n + 1];
			Edge[] edges = new Edge[n - 1];
			for (int i = 0; i < n - 1; ++i) {
				int u = scn.nextInt();
				int v = scn.nextInt();

				edges[i] = new Edge(i, u, v);

				if (tree[u] == null) {
					tree[u] = new ArrayList<>();
				}

				if (tree[v] == null) {
					tree[v] = new ArrayList<>();
				}

				tree[u].add(edges[i]);
				tree[v].add(edges[i]);
			}
			int m = scn.nextInt();
			long[] p = new long[m];
			for (int i = 0; i < m; ++i) {
				p[i] = scn.nextLong();
			}

			out.println(solve(n, tree, m, p, edges));
			t--;
		}
		out.close();
	}

	public static class Edge {
		int u;
		int v;
		int id;
		long c; // contribution in paths
		long wt;

		Edge(int id, int u, int v) {
			this.u = u;
			this.v = v;
			this.id = id;
		}
	}

	public static class eCom implements Comparator<Edge> {

		@Override
		public int compare(Edge arg0, Edge arg1) {
			// TODO Auto-generated method stub
			return Long.compare(arg0.c, arg1.c);
		}

	}

	public static long solve(int n, List<Edge>[] tree, int m, long[] p, Edge[] edges) {

		int[] sz = new int[n + 1];
		dfs(n, 1, 0, tree, sz);
		Arrays.sort(edges, new eCom());
		Arrays.sort(p);
		if (m > n - 1) {
			int i = m - 2;
			while (i >= n - 2) {
				p[i] *= p[i + 1];
				p[i] %= mod;
				i--;
			}

			for (int j = 0; j < n - 1; ++j) {
				edges[j].wt = p[j];
			}

		} else if (m <= n - 1) {

			int j = m - 1;
			int i = n - 2;
			while (j >= 0) {
				edges[i].wt = p[j];
				j--;
				i--;
			}

			while (i >= 0) {
				edges[i].wt = 1;
				i--;
			}

		}

		long ans = 0;

		for (int i = 0; i < n - 1; ++i) {
			long x = (edges[i].c * edges[i].wt) % mod;
			ans += x;
			ans %= mod;
		}
		return ans;
	}

	// subtree size including itself
	public static int dfs(int n, int root, int p, List<Edge>[] tree, int[] sz) {

		int ans = 1;
		for (Edge e : tree[root]) {
			int nbr = e.u == root ? e.v : e.u;
			if (nbr != p) {
				int szNbr = dfs(n, nbr, root, tree, sz);
				int rem = n - szNbr;
				e.c = (long) szNbr * rem;
				ans += szNbr;
			}
		}
		return sz[root] = ans;
	}

	static final long mod = (long) 1e9 + 7;

	static FastScanner scn = new FastScanner();
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));

	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

		String next() {
			while (!st.hasMoreTokens())
				try {
					st = new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}

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

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

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

}
