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.List;
import java.util.PriorityQueue;
import java.util.StringTokenizer;

public final class D {

    static class Node {
        List<Node> adj = new ArrayList<>();
        int val;

        Node(int val) {
            this.val = val;
        }
    }

    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) {
                    //noinspection CallToPrintStackTrace
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

        int[] readArray(int n) {
            final int[] a = new int[n];
            for (int i = 0; i < n; i++) { a[i] = nextInt(); }
            return a;
        }

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

    private static final long MOD = (long) (1e9 + 7);

    public static void main(String[] args) {
        final FastScanner fs = new FastScanner();
        final int t = fs.nextInt();
        for (int x = 0; x < t; x++) {
            final int n = fs.nextInt();
            final Node[] g = new Node[n];
            for (int i = 0; i < n; i++) {
                g[i] = new Node(i);
            }
            for (int i = 1; i < n; i++) {
                final int u = fs.nextInt() - 1;
                final int v = fs.nextInt() - 1;
                g[u].adj.add(g[v]);
                g[v].adj.add(g[u]);
            }
            final int m = fs.nextInt();
            final int[] primes = fs.readArray(m);
            final PriorityQueue<Integer> pq1 = new PriorityQueue<>();
            for (int i = 0; i < m; i++) { pq1.offer(primes[i]); }
            final long[] factors = new long[n - 1];
            final long[] size = new long[n];
            dfs(g[0], null, size);
            Arrays.fill(factors, 1);
            boolean changeLast = false;
            for (int i = 0; i < n - 1; i++) {
                factors[i] = pq1.isEmpty() ? 1 : pq1.remove();
                if (i == n - 2) {
                    while (!pq1.isEmpty()) {
                        factors[i] = (factors[i] * pq1.remove()) % MOD;
                        if (i > 0 && factors[i] < factors[i - 1]) {
                            changeLast = true;
                        }
                    }
                }
            }
            final long[] edgeWeights = new long[n - 1];
            // since we rooted the tree at node 0, size[0] * (n - size[0]) = 0;
            for (int i = 1; i < n; i++) {
                edgeWeights[i - 1] = size[i] * (n - size[i]);
            }
            long temp = 0;
            if (changeLast) {
                temp = factors[n - 2];
                factors[n - 2] = MOD;
            }
            Arrays.sort(edgeWeights);
            Arrays.sort(factors);
            if (changeLast) {
                factors[n - 2] = temp;
            }
            long ans = 0;
            for (int i = 0; i < n - 1; i++) {
                ans = (ans + ((factors[i] % MOD) * edgeWeights[i]) % MOD) % MOD;
            }
            System.out.println(ans);
        }
    }

    private static void dfs(Node curr, Node parent, long[] size) {
        size[curr.val] = 1;
        for (Node next : curr.adj) {
            if (next != parent) {
                dfs(next, curr, size);
                size[curr.val] += size[next.val];
            }
        }
    }

//    private static int subarraySum(int[] arr) {
//        int result = 0;
//        for (int i = 0; i < arr.length; i++) {
//            result += arr[i] * (i + 1) * (arr.length - i);
//        }
//        return result;
//    }
}
