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.BufferedOutputStream;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;

public class Contest665D {
	static ArrayList<Integer>[] adj;
	static int[] par;
	static long[] up, down;

	public static void main(String[] args) {
		MyScanner sc = new MyScanner();
		out = new PrintWriter(new BufferedOutputStream(System.out));
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			adj = new ArrayList[n+1];
			par = new int[n+1];
			up = new long[n+1];
			down = new long[n+1];
			for (int i = 0; i < adj.length; i++) {
				adj[i] = new ArrayList<Integer>();
			}
			for (int i = 0; i < n-1; i++) {
				int a = sc.nextInt();
				int b = sc.nextInt();
				adj[a].add(b);
				adj[b].add(a);
			}
			int m = sc.nextInt();
			int[] primes = new int[m];
			for (int i = 0; i < m; i++) {
				primes[i] = sc.nextInt();
			}
			Arrays.sort(primes);
			dfs(1);
			for (int i = 2; i <= n; i++) {
				down[i] = n - up[i];
			}
			long[] edge = new long[n+1];
			for (int i = 2; i <= n; i++) {
				edge[i] = (long) up[i] * down[i];
			}
			long ans = 0;
			Arrays.sort(edge);
			int MOD = 1000000007;
			if (m <= n-1) {
				int index = 0;
				while (index < m) {
					ans = ((long) ans + edge[edge.length-index-1] * primes[primes.length-index-1]) % MOD;
					index++;
				}
				for (int i = 2; i <= n-m; i++) {
					ans = ((long) ans + edge[i]) % MOD;
				}
				System.out.println(ans);
			}
			else {
				int diff = m-n+2;
				long mul = 1;
				for (int i = 0; i < diff; i++) {
					mul = ((long) mul* primes[primes.length-i-1]) % MOD;
				}
				ans = ((long) mul * edge[n]) % MOD;
				int index = 0;
				while (index < n-2) {
					ans = ((long) ans + edge[n-index-1] * primes[m-diff-1-index]) % MOD;
					index++;
				}
				System.out.println(ans);
			}
		}
		out.close();
	}
	
	public static long dfs(int x) {
		int sum = 1;
		for (int y: adj[x]) {
			if (y != par[x]) {
				par[y] = x;
				sum += dfs(y);
				
			}
		}
		return up[x] = sum;
	}
	
	public static void dfs2(int x) {
		int sum = 0;
		for (int y: adj[x]) {
			if (y != par[x]) {
				down[x] = up[par[x]] - up[x] + 1;
			}
		}
		down[x] = sum;
	}
	
	public static PrintWriter out;
	
	public static class MyScanner {
		BufferedReader br;
		StringTokenizer st;
		public MyScanner() {
			br = new BufferedReader(new InputStreamReader(System.in));
		}
		String next() {
			while (st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				}
				catch (IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}
		int nextInt() {
			return Integer.parseInt(next());
		}
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		String nextLine() {
			String str = "";
			try {
				str = br.readLine();
			}
			catch (IOException e) {
				e.printStackTrace();
			}
			return str;
		}
	}

}
