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 Main {

	static ArrayList<Integer>[] g;
	static boolean[] vis;

	public static void main(String[] args) throws NumberFormatException, IOException {
		// TODO Auto-generated method stub
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter pw = new PrintWriter(System.out);

		StringTokenizer st = new StringTokenizer(br.readLine());
		int t = Integer.parseInt(st.nextToken());
		while (t-- > 0) {
			n = Integer.parseInt(br.readLine());
			g = new ArrayList[n];
			for (int i = 0; i < g.length; i++) {
				g[i] = new ArrayList<>();
			}
			for (int i = 0; i < n - 1; i++) {
				st = new StringTokenizer(br.readLine());
				int u = Integer.parseInt(st.nextToken()) - 1;
				int v = Integer.parseInt(st.nextToken()) - 1;
				g[u].add(v);
				g[v].add(u);
			}
			int m = Integer.parseInt(br.readLine());
			int[] e = new int[n - 1];
			Arrays.fill(e, 1);
			st = new StringTokenizer(br.readLine());
			int[] arr = new int[m];
			for (int i = 0; i < arr.length; i++) {
				arr[i] = Integer.parseInt(st.nextToken());
			}
			Arrays.sort(arr);
			for (int i = 0; i < m; i++) {
				if (i < e.length - 1)
					e[i] = arr[i];
				else
					e[n - 2] = (int) ((e[n - 2] * 1l * arr[i]) % 1000000007);
			}
			temp = new ArrayList<>();
			dfs(0, -1);
			Collections.sort(temp);
			Arrays.sort(e);
			long sum = 0;
			for (int i = 0; i < e.length; i++) {
				sum += (temp.get(i) % 1000000007) * e[i];
				sum %= 1000000007;
			}
			pw.println(sum);
		}

		pw.flush();
		pw.close();

	}

	static int n;
	static ArrayList<Long> temp;

	static int dfs(int u, int p) {
		int sum = 1;
		for (int x : g[u]) {
			if (x != p) {
				int t = dfs(x, u);
				temp.add(t * 1l * (n - t));
				sum += t;
			}
		}
		return sum;
	}

	static ArrayList<Integer> primes;
	static int[] isComposite;

	static void sieve(int N) // O(N log log N)
	{
		isComposite = new int[N + 1];
		isComposite[0] = isComposite[1] = 1; // 0 indicates a prime number
		primes = new ArrayList<Integer>();

		for (int i = 2; i <= N; ++i) // can loop till i*i <= N if primes array is not needed O(N log log sqrt(N))
			if (isComposite[i] == 0) // can loop in 2 and odd integers for slightly better performance
			{
				primes.add(i);
				if (1l * i * i <= N)
					for (int j = i * i; j <= N; j += i) // j = i * 2 will not affect performance too much, may alter in
														// modified sieve
						isComposite[j] = 1;
			}
	}

	static long pow(long a, long e, int mod) // O(log e)
	{
		long res = 1;
		while (e > 0) {
			if ((e & 1) == 1) {
				res *= a;
				res %= mod;
			}
			a *= a;
			a %= mod;
			e >>= 1;
		}
		return res;
	}

	static int maxSize;
	static boolean[][] exist;
	static ArrayList<Integer>[] startAt, endAt;

	static void compress(pair[] arr) {
		TreeSet<Integer> tr = new TreeSet<>();
		for (int i = 0; i < arr.length; i++) {
			tr.add(arr[i].x);
			tr.add(arr[i].y);
		}
		tr.add(0);
		tr.add((int) 1e9);
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : tr)
			hm.put(x, hm.size());
		for (int i = 0; i < arr.length; i++) {
			arr[i].x = hm.get(arr[i].x);
			arr[i].y = hm.get(arr[i].y);
		}
		maxSize = hm.size();
	}

	static long gcd(long a, long b) {
		if (a == 0)
			return b;
		return gcd(b % a, a);
	}

	static class pair implements Comparable<pair> {
		int x;
		int y;

		public pair(int d, int u) {
			x = d;
			y = u;
		}

		@Override
		public int compareTo(pair o) {
			// TODO Auto-generated method stub
			int x1 = y - x;
			int x2 = o.y - o.x;
			return x1 - x2;
		}

	}

}