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.*;
import java.util.*;

public class D {

	static long times[];
	static ArrayList<Integer> adj[];
	static int n;
	static boolean vis[];

	static int dfs(int u) {

		vis[u] = true;
		int t = 1;
		for (int v : adj[u]) {
			if (!vis[v])
				t += dfs(v);
		}
		return (int) (times[u] = t);
	}

	public static void main(String[] args) throws Exception {
		Scanner sc = new Scanner(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			n = sc.nextInt();
			adj = new ArrayList[n];
			vis = new boolean[n];
			times = new long[n];
			for (int i = 0; i < n; i++) {
				adj[i] = new ArrayList<>();
			}
			for (int i = 0; i < n - 1; i++) {
				int u = sc.nextInt() - 1;
				int v = sc.nextInt() - 1;
				adj[u].add(v);
				adj[v].add(u);
			}
			int k = sc.nextInt();
			ArrayList<Integer> p = new ArrayList<>();
			for (int i = 0; i < k; i++) {
				p.add(sc.nextInt());
			}
			dfs(0);

			for (int i = 0; i < n; i++) {
				times[i] *= (n - times[i]);
			}
			shuffle(times);
			Arrays.sort(times);
			Collections.sort(p);
			int x = 0;
			long ans = 0;
			int mod = (int) 1e9 + 7;
			long z[] = new long[n];
			Arrays.fill(z, 1);
			int g = Math.max(1, n - 1 - k + 1);
			// System.out.println(g);
			int u = 0;
			for (int i = g; i < z.length; i++) {
				z[i] = p.get(u++);
			}
			if (u < k) {
				for (int i = u; i < k; i++) {
					z[n - 1] = (z[n - 1] * p.get(i)) % mod;
				}
			}
			// System.out.println(Arrays.toString(times));
			// System.out.println(Arrays.toString(z));
			for (int i = n - 1; i > 0; i--) {
				ans = (ans + (z[i] * times[i]) % mod) % mod;
			}
			System.out.println(ans);
		}
	}

	// int x=32+16+4+2+2+8+1
	///////////////////////////////////////////////////////////////////////////////////////////
	static class tri {
		int x, y, z;

		public tri(int x, int y, int z) {
			this.x = x;
			this.y = y;
			this.z = z;
		}

	}

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

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

		@Override
		public int compareTo(pair p) {
			return (x == p.x && y == p.y) ? 0 : 1;
		}

		@Override
		public String toString() {
			return x + " " + y;

		}

	}

	static long mod(long ans, int mod) {
		return (ans % mod + mod) % mod;
	}

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

	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	public static int log(int n, int base) {
		int ans = 0;
		while (n + 1 > base) {
			ans++;
			n /= base;
		}
		return ans;
	}

	static long pow(long b, long e) {
		long ans = 1;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = ((ans * 1l * b));
			e >>= 1;
			{

			}
			b = ((b * 1l * b));
		}
		return ans;
	}

	static long powmod(long r, long e, int mod) {
		long ans = 1;
		r %= mod;
		while (e > 0) {
			if ((e & 1) == 1)
				ans = (int) ((ans * 1l * r) % mod);
			e >>= 1;
			r = (int) ((r * 1l * r) % mod);
		}
		return ans;
	}

	static int ceil(int a, int b) {
		int ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	static long ceil(long a, long b) {
		long ans = a / b;
		return a % b == 0 ? ans : ans + 1;
	}

	static HashMap<Integer, Integer> compress(int a[]) {
		TreeSet<Integer> ts = new TreeSet<>();
		HashMap<Integer, Integer> hm = new HashMap<>();
		for (int x : a)
			ts.add(x);
		for (int x : ts) {
			hm.put(x, hm.size() + 1);
		}
		return hm;
	}

	// Returns nCr % p
	static int C[];

	static int nCrModp(int n, int r, int p) {
		if (r > n - r)
			r = n - r;
		if (C[r] != 0)
			return C[r];
		// The array C is going to store last
		// row of pascal triangle at the end.
		// And last entry of last row is nCr

		C[0] = 1; // Top row of Pascal Triangle

		// One by constructs remaining rows of Pascal
		// Triangle from top to bottom
		for (int i = 1; i <= n; i++) {

			// Fill entries of current row using previous
			// row values
			for (int j = Math.min(i, r); j > 0; j--)

				// nCj = (n-1)Cj + (n-1)C(j-1);
				C[j] = (C[j] + C[j - 1]) % p;
		}
		return C[r];
	}

	static class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

		public Scanner(String file) throws Exception {
			br = new BufferedReader(new FileReader(file));
		}

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

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

		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 String nextLine() throws IOException {
			return br.readLine();
		}

		public double nextDouble() throws IOException {
			String x = next();
			StringBuilder sb = new StringBuilder("0");
			double res = 0, f = 1;
			boolean dec = false, neg = false;
			int start = 0;
			if (x.charAt(0) == '-') {
				neg = true;
				start++;
			}
			for (int i = start; i < x.length(); i++)
				if (x.charAt(i) == '.') {
					res = Long.parseLong(sb.toString());
					sb = new StringBuilder("0");
					dec = true;
				} else {
					sb.append(x.charAt(i));
					if (dec)
						f *= 10;
				}
			res += Long.parseLong(sb.toString()) / f;
			return res * (neg ? -1 : 1);
		}

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

	}

	public static void shuffle(long[] times2) {
		int n = times2.length;
		for (int i = 0; i < n; i++) {
			int r = i + (int) (Math.random() * (n - i));
			long tmp = times2[i];
			times2[i] = times2[r];
			times2[r] = tmp;
		}
	}

}