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 D_MaximumDistributedTree {
    static final int MOD = 1_000_000_007;

    private static final BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    private static final PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
    private static StringTokenizer st;

    private static int readInt() throws IOException {
        while (st == null || !st.hasMoreTokens()) st = new StringTokenizer(br.readLine());
        return Integer.parseInt(st.nextToken());
    }

    public static void main(String[] args) throws IOException {
        int T = readInt();
        while (T-- > 0) pw.println(solve());
        pw.close();
    }

    private static long solve() throws IOException {
        int n = readInt();
        ArrayList<Integer>[] adj = new ArrayList[n + 1];
        for (int i = 1; i < n; i++) {
            int a = readInt();
            int b = readInt();
            if (adj[a] == null) adj[a] = new ArrayList<>();
            adj[a].add(b);
            if (adj[b] == null) adj[b] = new ArrayList<>();
            adj[b].add(a);
        }

        int[] parent = new int[n + 1];
        final int ROOT = 1;
        final int PARENT_OF_ROOT = 0;
        parent[ROOT] = PARENT_OF_ROOT;

        // breath-first-traversal starting from 1
        int[] bFS = new int[n];
        int queueTail = 0;
        bFS[queueTail] = 1;

        int queueHead = -1;
        while (++queueHead < n) {
            int p = bFS[queueHead];
            for (int c : adj[p]) {
                if (c != parent[p]) {
                    bFS[++queueTail] = c;
                    parent[c] = p;
                }
            }
        }

        ArrayList<Long> rep = new ArrayList<>(n - 1);
        long[] subtreeNodes = new long[n + 1];
        // reversed breath-first-traversal
        for (int i = n - 1; i >= 1; i--) {
            int j = bFS[i];
            subtreeNodes[j]++;
            rep.add(subtreeNodes[j] * (n - subtreeNodes[j]));
            subtreeNodes[parent[j]] += subtreeNodes[j];
        }

        Collections.sort(rep);

        int m = readInt();
        long[] primes = new long[m];
        for (int i = 0; i < m; i++) {
            primes[i] = readInt();
        }
        Arrays.sort(primes);

        long[] multiplier = new long[n - 1];
        if (m <= n - 1) {
            Arrays.fill(multiplier, 0, n - 1 - m, 1);
            System.arraycopy(primes, 0, multiplier, n - 1 - m, m);
        } else {
            System.arraycopy(primes, 0, multiplier, 0, n - 1);
            for (int i = n - 1; i < m; i++) {
                multiplier[n - 2] = multiplier[n - 2] * primes[i] % MOD;
            }
        }

        long res = 0;
        for (int i = 0; i < n - 1; i++) {
            res = (res + rep.get(i) * multiplier[i] % MOD) % MOD;
        }

        return res;
    }
}
