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.math.*;
import java.util.*;
 
public class Main {
	static final long MOD = 1000000007L;
	static final int INF = 50000000;
	
	static ArrayList<Integer>[] tree;
	static int[] children;
	static int[] parent;
	public static void main(String[] args) {
		FastScanner sc = new FastScanner();
		PrintWriter pw = new PrintWriter(System.out);
		
		int Q = sc.ni();
		for (int q = 0; q < Q; q++) {
			int N = sc.ni();
			tree = new ArrayList[N];
			for (int i  =0; i < N; i++)
				tree[i] = new ArrayList<Integer>();
			for (int i =0; i < N-1; i++) {
				int n1 = sc.ni()-1;
				int n2 = sc.ni()-1;
				tree[n1].add(n2);
				tree[n2].add(n1);
			}
			children = new int[N];
			parent = new int[N];
			dfs(0);
			
			ArrayList<Long> paths = new ArrayList<Long>();
			for (int i = 0; i < N; i++) {
				for (int j: tree[i]) {
					if (j==parent[i])continue;
					paths.add((N+0L-children[j])*children[j]);
				}
			}
			Collections.sort(paths);
			
			int P = sc.ni();
			ArrayList<Long> factors = new ArrayList<Long>();
			for (int i = 0; i < P; i++) {
				factors.add(sc.nl());
			}
			Collections.sort(factors);
			if (P <= N-1) {
				for (int i = P; i < N-1; i++) {
					factors.add(1L);
				}
				Collections.sort(factors);
			} else {
				ArrayList<Long> combine = new ArrayList<Long>();
				for (int i = 0; i < N-2; i++) {
					combine.add(factors.get(i));
				}
				long v = 1;
				for (int i = N-2; i < P; i++) {
					v = (v*factors.get(i))%MOD;
				}
				combine.add(v);
				factors = combine;
			}
			
			long ans = 0;
			for (int i = 0; i < N-1; i++) {
				long n1 = paths.get(i)%MOD;
				long n2 = factors.get(i)%MOD;
				long n = (n1*n2)%MOD;
				ans = (ans+n)%MOD;
			}
			pw.println(ans);
		}
		pw.close();
	}
	
	public static int dfs(int node) {
		children[node] += 1;
		for (int child: tree[node]) {
			if (child != parent[node]) {
				parent[child] = node;
				children[node] += dfs(child);
			}
		}
		return children[node];
	}

	
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            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 ni() { 
            return Integer.parseInt(next()); 
        }
  
        long nl() { 
            return Long.parseLong(next()); 
        } 
  
        double nd() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() {
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str;
        }
    }
}