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 MainClass
{
    static long M = (long)1e9 + 7;
    static ArrayList<Long> edges = new ArrayList<>();
    static int n;
    public static void main(String args[])throws IOException
    {
        Reader in = new Reader();
        int t = in.nextInt();
        StringBuilder stringBuilder = new StringBuilder();
        while (t-- > 0)
        {
            n = in.nextInt();
            ArrayList<Integer>[] adj = new ArrayList[n];
            for (int i=0;i<n;i++)
                adj[i] = new ArrayList<>();
            for (int i=0;i<n - 1;i++)
            {
                int x = in.nextInt() - 1;
                int y = in.nextInt() - 1;
                adj[x].add(y);
                adj[y].add(x);
            }
            edges.clear();
            int m = in.nextInt();
            ArrayList<Long> primes = new ArrayList<>();
            for (int i=0;i<m;i++)
                primes.add(in.nextLong());
            if (m < n - 1)
            {
                for (int i=m;i<n - 1;i++)
                    primes.add(1L);
            }
            Collections.sort(primes);
            explore(adj, 0, new boolean[n]);
            Collections.sort(edges);
            long ans = 0L;
            for (int i=0;i<n - 2;i++)
            {
                ans = (ans + (edges.get(i) * primes.get(i)) % M) % M;
            }
            long ll = 1L;
            for (int i=n - 2;i<primes.size();i++)
            {
                ll = (ll * primes.get(i)) % M;
            }
            ans = (ans + (ll * edges.get(n - 2)) % M) % M;

            stringBuilder.append(ans).append("\n");
        }
        System.out.println(stringBuilder);
    }
    public static long explore(ArrayList<Integer>[] adj, int v, boolean[] visited)
    {
        visited[v] = true;
        long ans = 1L;
        for (int u: adj[v])
        {
            if (!visited[u])
            {
                ans += explore(adj, u, visited);
            }
        }
        if (v != 0)
            edges.add(ans * (n - ans));
        return ans;
    }
}
class Reader
{
    final private int BUFFER_SIZE = 1 << 16;
    private DataInputStream din;
    private byte[] buffer;
    private int bufferPointer, bytesRead;

    public Reader()
    {
        din = new DataInputStream(System.in);
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public Reader(String file_name) throws IOException
    {
        din = new DataInputStream(new FileInputStream(file_name));
        buffer = new byte[BUFFER_SIZE];
        bufferPointer = bytesRead = 0;
    }

    public String readLine() throws IOException
    {
        byte[] buf = new byte[64]; // line length
        int cnt = 0, c;
        while ((c = read()) != -1)
        {
            if (c == '\n')
                break;
            buf[cnt++] = (byte) c;
        }
        return new String(buf, 0, cnt);
    }

    public int nextInt() throws IOException
    {
        int ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do
        {
            ret = ret * 10 + c - '0';
        }  while ((c = read()) >= '0' && c <= '9');

        if (neg)
            return -ret;
        return ret;
    }

    public long nextLong() throws IOException
    {
        long ret = 0;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();
        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');
        if (neg)
            return -ret;
        return ret;
    }

    public double nextDouble() throws IOException
    {
        double ret = 0, div = 1;
        byte c = read();
        while (c <= ' ')
            c = read();
        boolean neg = (c == '-');
        if (neg)
            c = read();

        do {
            ret = ret * 10 + c - '0';
        }
        while ((c = read()) >= '0' && c <= '9');

        if (c == '.')
        {
            while ((c = read()) >= '0' && c <= '9')
            {
                ret += (c - '0') / (div *= 10);
            }
        }

        if (neg)
            return -ret;
        return ret;
    }

    private void fillBuffer() throws IOException
    {
        bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
        if (bytesRead == -1)
            buffer[0] = -1;
    }

    private byte read() throws IOException
    {
        if (bufferPointer == bytesRead)
            fillBuffer();
        return buffer[bufferPointer++];
    }

    public void close() throws IOException
    {
        if (din == null)
            return;
        din.close();
    }
}
