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 A {
	static BufferedReader br;
	static PrintWriter pw;
	static int inf = (int) 1e9;
	static long mod = (long) 1e9 + 7;

	static ArrayList<Integer>[] graph;
	static int n;

	public static void main(String[] args) throws NumberFormatException, IOException, InterruptedException {
//		br = new BufferedReader(new FileReader(new File("chief.in")));
//		br = new BufferedReader(new InputStreamReader(System.in));
		Scanner sc = new Scanner(System.in);
		pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			n = sc.nextInt();
			graph = new ArrayList[n];
			for (int i = 0; i < graph.length; i++) {
				graph[i] = new ArrayList<>();
			}
			for (int i = 0; i < n - 1; i++) {
				int u = sc.nextInt() - 1;
				int v = sc.nextInt() - 1;
				graph[u].add(v);
				graph[v].add(u);
			}
			int p = sc.nextInt();
			ArrayList<Integer> arr = new ArrayList<>();
			for (int i = 0; i < p; i++) {
				arr.add(sc.nextInt());
			}
			for (int i = 0; i < n - 1 - p; i++) {
				arr.add(1);
			}
			ans = new ArrayList<>();
			dfs(0, -1);
			Collections.sort(arr, Collections.reverseOrder());
			Collections.sort(ans, Collections.reverseOrder());
			if (p >= n) {
				long first = 1;
				for (int i = 0; i < p - n + 2; i++) {
					first *= arr.get(i);
					first %= mod;
				}
				ArrayList<Integer> arr2 = new ArrayList<>();
				arr2.add((int) first);
				for (int i = p - n + 2; i < p; i++) {
					arr2.add(arr.get(i));
				}
				long res = 0;
				for (int i = 0; i < n - 1; i++) {
					res += ans.get(i) * arr2.get(i);
					res %= mod;
				}
				pw.println(res);
			} else {
				long res = 0;
				for (int i = 0; i < n - 1; i++) {
					res += ans.get(i) * arr.get(i);
					res %= mod;
				}
				pw.println(res);
			}
		}
		pw.flush();
		pw.close();

	}

	static long[][] dp;
	static ArrayList<Long> ans;

	static int dfs(int u, int pa) {
		int sum = 1;
		for (int v : graph[u]) {
			if (v == pa)
				continue;
			int x = dfs(v, u);
			ans.add(x * 1l * (n - x));
			sum += x;
		}
		return sum;
	}

	static long[] f, g;

	static void numOfSubtrees(int u, int pa) {
		f[u] = 1;
		for (int v : graph[u]) {
			if (v == pa)
				continue;
			numOfSubtrees(v, u);
			f[u] *= (1 + f[v]);
			g[u] += f[v] + g[v];
		}
	}

	static ArrayList<Integer> primes; // generated by sieve

	/*
	 * 1. Generating a list of prime factors of N
	 */
	static ArrayList<Long> primeFactors(Long N) // O(sqrt(N) / ln sqrt(N))
	{
		ArrayList<Long> factors = new ArrayList<Long>(); // take abs(N) in case of -ve integers
		int idx = 0, p = primes.get(idx);

		while (p * p <= N) {
			if (N % p == 0)
				factors.add((long) p);
			while (N % p == 0) {
				N /= p;
			}
			if (idx + 1 == primes.size())
				break;
			p = primes.get(++idx);
		}

		if (N != 1) // last prime factor may be > sqrt(N)
			factors.add(N); // for integers whose largest prime factor has a power of 1
		return factors;
	}

	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 class Scanner {
		StringTokenizer st;
		BufferedReader br;

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

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

	}

	static int m, k;

	static double[] memo;
	static ArrayList<Integer>[] adj;

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

	// method to return LCM of two numbers
	static long lcm(long a, long b) {
		return (a * b) / gcd(a, b);
	}

	static double dis(int x, int y, int z, int w) {
		return Math.sqrt((x - z) * (x - z) + (y - w) * (y - w));
	}

	static int[] nxtarr() throws IOException {
		StringTokenizer st = new StringTokenizer(br.readLine());
		int[] a = new int[st.countTokens()];
		for (int i = 0; i < a.length; i++) {
			a[i] = Integer.parseInt(st.nextToken());
		}
		return a;
	}

	static long[] nxtarrLong() throws IOException {
		StringTokenizer st = new StringTokenizer(br.readLine());
		long[] a = new long[st.countTokens()];
		for (int i = 0; i < a.length; i++) {
			a[i] = Long.parseLong(st.nextToken());
		}
		return 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
			if (y == o.y)
				return x - o.x;
			return y - o.y;
		}

	}

	static class triple implements Comparable<triple> {
		int x;
		int y;
		int z;

		public triple(int a, int b, int c) {
			x = a;
			y = b;
			z = c;
		}

		@Override
		public int compareTo(triple o) {
			// TODO Auto-generated method stub
			return x - o.x;
		}
	}

}
