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 D1401 {

	public static void main(String[] args) throws IOException {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			ArrayList<Integer>[] adjList = new ArrayList[n];
			for (int i = 0; i < adjList.length; i++) {
				adjList[i] = new ArrayList<Integer>();
			}
			for (int i = 0; i < n - 1; i++) {
				int u = sc.nextInt() - 1;
				int v = sc.nextInt() - 1;
				adjList[u].add(v);
				adjList[v].add(u);
			}
			int m = sc.nextInt();
			Long[] prime = new Long[m];
			for (int i = 0; i < prime.length; i++) {
				prime[i] = sc.nextLong();
			}
			Arrays.sort(prime);
			int[] p = new int[n];
			int[] deg = new int[n];
			Queue<Integer> leaves = new LinkedList<Integer>();
			for (int i = 0; i < deg.length; i++) {
				deg[i] = adjList[i].size();
				if (deg[i] == 1 && i != 0)
					leaves.add(i);
			}
			Queue<Integer> q = new LinkedList<Integer>();
			p[0] = -1;
			q.add(0);
			while (!q.isEmpty()) {
				int cur = q.poll();
				for (int x : adjList[cur]) {
					if (x != p[cur]) {
						q.add(x);
						p[x] = cur;
					}
				}
			}
			deg[0]++;
			int[] sub = new int[n];
			while (!leaves.isEmpty()) {
				int cur = leaves.poll();
				sub[cur]++;
				if (cur == 0)
					break;
				sub[p[cur]] += sub[cur];
				if (--deg[p[cur]] == 1) {
					leaves.add(p[cur]);
				}
			}
			ArrayList<Long> al = new ArrayList<Long>();
			for (int i = 0; i < n; i++) {
				for (int x : adjList[i]) {
					if (p[x] == i) {
//						pw.println(i + " " + x + " " + (1l * sub[x] * (n - sub[x])));
						al.add(1l * sub[x] * (n - sub[x]));
					}
				}
			}
//			pw.println(Arrays.toString(sub));
			Collections.sort(al);
			int mod = (int) 1e9 + 7;
			long ans = 0;
			int idxedge = al.size() - 1;
			int idxprime = m - 1;
//			pw.println(al.size());
//			pw.println(al);
//			pw.println(Arrays.toString(prime));
			while (idxprime > idxedge) {
				prime[idxprime - 1] = (prime[idxprime - 1] * prime[idxprime]) % mod;
				idxprime--;
			}
			while (idxedge >= 0) {
				if (idxprime == -1) {
					ans += al.get(idxedge--) % mod;
				} else {
					ans += (1l * al.get(idxedge--) * prime[idxprime--]) % mod;
				}
				ans %= mod;
			}
			pw.println(ans);
		}
		pw.close();

	}

	static class Scanner {
		BufferedReader br;
		StringTokenizer st;

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

		public Scanner(FileReader f) {
			br = new BufferedReader(f);
		}

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

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

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

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

		public int[] nextIntArr(int n) throws IOException {
			int[] arr = new int[n];
			for (int i = 0; i < n; i++) {
				arr[i] = Integer.parseInt(next());
			}
			return arr;
		}

	}

}
