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 FastReader f = new FastReader();
	static PrintWriter out = new PrintWriter(System.out);

	public static void main(String[] args) {
		int t = f.nextInt();
		while(t-- > 0) {
			solve();
		}
		out.close();
	}

	static int MOD = (int) (1e9+7);
	static Node[] nodes;
	static int n;
	static void solve() {
		n = f.nextInt();
		nodes = new Node[n];
		ans = new ArrayList<>();
		for(int i=0;i<n;i++) {
			nodes[i] = new Node();
		}
		for(int i=1;i<n;i++) {
			int u = f.nextInt() - 1, v = f.nextInt() - 1;
			nodes[u].adjacent.add(v);
			nodes[v].adjacent.add(u);
		}


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


		Arrays.sort(primes);
		if(n <= m) {
			for(int i=n-1;i<m;i++) {
				primes[n-2] = (int) ((((long)primes[n-2]) * primes[i]) % MOD);  
			}
		}

		dfs(-1,0);
		dfs2(-1,0);
		Collections.sort(ans);
		int pointPrime = Math.min(n-2,m-1);
		long an = 0;
		for(int i=ans.size()-1;i>=0;i--) {
			long mul = 1;
			if(pointPrime >= 0) {
				mul = primes[pointPrime];
				pointPrime--;
			}

			an = (an + mul * ans.get(i)) % MOD;
		}


		out.println(an);
	}
	static ArrayList<Long> ans;

	static void dfs2(int par, int now) {

		for(int i : nodes[now].adjacent) {
			if(i != par) {
				ans.add(((n-1 + ((long)nodes[i].subtreeSize) * (n-2-nodes[i].subtreeSize))));
				dfs2(now, i);
			}
		}
	}

	static int dfs(int par, int now) {
		int subSize = 0;
		for(int i : nodes[now].adjacent) {
			if(i != par) {
				subSize += dfs(now, i);
			}
		}

		nodes[now].subtreeSize = subSize; 
		return subSize + 1;
	}



	static class Node {
		int subtreeSize;
		ArrayList<Integer> adjacent = new ArrayList<>();
	}

	





	static class FastReader {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st;

		String next() {
			while(st == null || !st.hasMoreElements()) {
				try {
					st = new StringTokenizer(br.readLine());
				} catch(IOException e) {
					e.printStackTrace();
				}
			}
			return st.nextToken();
		}

		String nextLine() {
			String str = "";

			try {
				str = br.readLine();
			} catch(IOException e) {
				e.printStackTrace();
			}
			return str;
		}

		int nextInt() {
			return Integer.parseInt(next());
		}

		double nextDouble() {
			return Double.parseDouble(next());
		}

		long nextLong() {
			return Long.parseLong(next());
		}
	}
}