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.time.LocalDate;


public class example {
	static HashSet<Integer>[] adj;
	static boolean[] vis;
	static long curSum;
	static ArrayList<Long> weights;
	static int n;
	public static void main(String[] args) throws Exception {
		int t = i();
		int MOD = 1_000_000_007;
		for (int tt = 0; tt < t; tt++) {
			n = i();
			adj = new HashSet[n];
			for (int i = 0; i < n; i++) {
				adj[i] = new HashSet<>();			
			}
			for (int i = 0; i < n - 1; i++) {
				int u = i() - 1, v = i() - 1;
				adj[u].add(v);
				adj[v].add(u);
			}

			vis = new boolean[n];
			vis[0] = true;
			weights = new ArrayList<>();
			dfs(0);
			Collections.sort(weights);

			int m = i();

			int[] primes = new int[m];
			for (int i = 0; i < m; i++) {
				primes[i] = i();
			}
			Arrays.sort(primes);



			long ans = 0;
			int primes_i = m - 1;
			if (m >= weights.size()) {
				long prod = 1;
				for (int i = m - 1; i >= weights.size() - 1; i--) {
					prod = prod * primes[i] % MOD;
				}
				ans = (ans + prod * weights.get(weights.size() - 1)) % MOD;
				primes_i = weights.size() - 2;
			} else {
				ans = (ans + weights.get(weights.size() - 1) * primes[m - 1]) % MOD;
				primes_i = m - 2;
			}
			for (int i = weights.size() - 2; i >= 0; i--) {
				if (primes_i < 0) {
					ans = (ans + weights.get(i)) % MOD;
				} else {
					ans = (ans + weights.get(i) * primes[primes_i--]) % MOD;
				}
			}

			out.println(ans);
		}

		out.close();
	}

	static int dfs(int node) {
		int sum = 0;
		for (Integer v : adj[node]) {
			if (!vis[v]) {
				vis[v] = true;
				int sub = dfs(v);
				weights.add((long) sub * (n - sub));
				sum += sub;
			}
		}
		return sum + 1;
	}


	static int r(int lo, int hi) {
		return (int) (Math.random() * (hi - lo + 1)) + lo;
	}


	static BufferedReader in;
	static {
		try {
			in = new BufferedReader(new FileReader("test.in"));
		} catch (Exception e) {
			in = new BufferedReader(new InputStreamReader(System.in));
		}
	}
	static PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)), true);
	static StringTokenizer st = new StringTokenizer("");
	static void check() throws Exception {
		while (!st.hasMoreTokens()) st = new StringTokenizer(in.readLine());
	}
	static int i() throws Exception {check(); return Integer.parseInt(st.nextToken());}
	static String s() throws Exception {check(); return st.nextToken();}
	static double d() throws Exception {check(); return Double.parseDouble(st.nextToken());}
	static long l() throws Exception {check(); return Long.parseLong(st.nextToken());}
}