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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Random;
import java.util.StringTokenizer;

public class Main {
 
	static long mod=(long)1e9+7;
	
	public static void main(String[] args) {
		FastScanner fs = new FastScanner();
		int tc = fs.nextInt();
		for(int t = 0;t < tc;t++) {
			int n = fs.nextInt();
			Nodes[] nodes = new Nodes[n];
			ArrayList<Long> s = new ArrayList<>();
			ArrayList<Long> c = new ArrayList<>();
			for(int i = 0;i < n;i++) {
				nodes[i] = new Nodes();
			}
			for(int i = 1;i < n;i++) {
				int fv = fs.nextInt() - 1;
				int sv = fs.nextInt() - 1;
				nodes[fv].list.add(sv);
				nodes[sv].list.add(fv);
			}
			int m = fs.nextInt();
			for(int i = 0;i < m;i++) {
				c.add(fs.nextLong());
			}
			dfs(nodes , 0 , 0 , s);
			Collections.sort(s);
			Collections.sort(c);
			
			for(int i = 0;i < m;i++) {
				if(i >= n - 1) {
					long cur = ((c.get(n - 2) % mod) * c.get(i) % mod) % mod;
					c.set(n - 2, cur);
				}
			}
			long ans = 0L;
			int cnt = Math.min(s.size(), c.size()) - 1;
			for(int i = s.size() - 1;i >= 0;i--) {
				long cur = ((cnt >= 0) ? c.get(cnt) : 1L);
				cnt--;
				long add = ((s.get(i) % mod )* (cur % mod)) % mod;
				ans = ((ans % mod) + (add % mod))% mod;
			}
			System.out.println(ans);
		}
	}
	
	public static long dfs(Nodes[] nodes , int root , int parent , ArrayList<Long> s) {
		
		long sum = 1l;
		for(int i = 0;i < nodes[root].list.size();i++) {
			int nxt = nodes[root].list.get(i);
			if(nxt == parent) 
				continue;
			long temp = dfs(nodes , nxt , root , s);
			s.add(temp*(nodes.length - temp));
			sum += temp;
		}
		return sum;
	}
	
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}

class Nodes {
	ArrayList<Integer> list = new ArrayList<Integer>();
}


