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.*;

public class C {
    static ArrayList<Integer> map[];
    static int mod = 1000000007;

    static int n;

    static LinkedList<Long> values;
    static boolean vis[];

    static long dfs(int node, int par) {
        long ans = 1;
        vis[node] = true;
        for (int a : map[node]) {
            if (a != par && !vis[a]) {
                long evaluated = dfs(a, node);
                values.add(evaluated * (int) (n - evaluated));
                ans += evaluated;
            }
        }

        return ans;
    }

    static long modulo(long x) {
        return (x % mod + mod) % mod;
    }

    static long mul(long a, long b) {
        return modulo(modulo(a) * modulo(b));
    }

    static long add(long a, long b) {
        return modulo(modulo(a) + modulo(b));
    }

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int testCases = sc.nextInt();
        StringBuilder sb = new StringBuilder();
        for (int t = 0; t < testCases; t++) {

            n = sc.nextInt();

            map = new ArrayList[n + 1];

            values = new LinkedList<>();
            vis = new boolean[n + 1];

            for (int i = 0; i <= n; i++) {
                map[i] = new ArrayList<>();
            }

            for (int i = 0; i < n - 1; i++) {
                int start = sc.nextInt();
                int end = sc.nextInt();
                map[start].add(end);
                map[end].add(start);
            }

            LinkedList<Long> primeFactor = new LinkedList<>();
            int pf = sc.nextInt();
            for (int i = 0; i < pf; i++) {
                int cur = sc.nextInt();
                primeFactor.add((long) cur);
            }

            Collections.sort(primeFactor, Collections.reverseOrder());

            if (primeFactor.size() < n - 1) {
                while (primeFactor.size() != n - 1) {
                    primeFactor.add((long) 1);
                }
            } else if (primeFactor.size() > n - 1) {
                long remove = 1;
                while (primeFactor.size() != n - 2) {
                    remove = mul(remove, primeFactor.remove(0));
                }
                primeFactor.add(0, remove);
            }

            dfs(1, -1);

            Collections.sort(values, Collections.reverseOrder());

            long ans = 0;
            while (!values.isEmpty() && !primeFactor.isEmpty()) {
                ans = add(ans, mul(values.remove(0), primeFactor.remove(0)));
            }
            sb.append(ans).append("\n");
        }
        System.out.print(sb);
    }
}