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.PrintWriter;
import java.util.*;

public class MaximumDistributedTree {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        for (int i = 0; i < t; i++) {
            solve(sc, pw);
        }
        pw.close();
    }
    static int mod = (int) 1e9 + 7;
    static List<int[]> ls = new ArrayList<>();
    static void solve(Scanner in, PrintWriter out){
        int n = in.nextInt();
        int[][] arr = new int[n - 1][];
        Map<Integer, List<Integer>> mp = new HashMap<>();
        for (int i = 0; i < n - 1; i++) {
            arr[i] = new int[]{in.nextInt() - 1, in.nextInt() - 1};
            if (!mp.containsKey(arr[i][0])) mp.put(arr[i][0], new ArrayList<>());
            if (!mp.containsKey(arr[i][1])) mp.put(arr[i][1], new ArrayList<>());
            mp.get(arr[i][0]).add(arr[i][1]);
            mp.get(arr[i][1]).add(arr[i][0]);
        }
        int m = in.nextInt();
        int[] prr = new int[m];
        for (int i = 0; i < m; i++) {
            prr[i] = in.nextInt();
        }
        dfs(n, 0, mp, -1);
        long res = 0;
        Collections.sort(ls, (a, b) -> {
            int dif1 = Math.abs(a[0] - a[1]);
            int dif2 = Math.abs(b[0] - b[1]);
            return dif1 - dif2;
        });
        Arrays.sort(prr);
//        System.out.println(Arrays.toString(prr));
        if (m > n - 1){
            int[] fk = ls.get(0);
//            System.out.println(Arrays.toString(fk));
            long pp = fk[0] * 1l * fk[1];
            pp %= mod;
            long shit = 1l;
            while (m > n - 2){
                shit = (shit * prr[m - 1]) % mod;
                m--;
            }
            res = (res + (pp * shit) % mod) % mod;
            for (int i = 1; i < ls.size(); i++, m--) {
                fk = ls.get(i);
//                System.out.println(Arrays.toString(fk));
//                System.out.println(prr[m - 1]+" prrval: ");
                pp = fk[0] * 1l * fk[1];
                pp %= mod;
                res = (res + (pp * prr[m - 1]) % mod) % mod;
            }
        }else{
            for (int i = 0; i < ls.size(); i++, m--) {
                int[] fk = ls.get(i);
//                System.out.println(Arrays.toString(fk));
                long pp = fk[0] * 1l * fk[1];
                pp %= mod;
                if (m > 0){
                    res = (res + (pp * prr[m - 1]) % mod) % mod;
                }else{
                    res = (res + pp) % mod;
                }
            }
        }
//        System.out.println(res+" fk:");
        out.println(res);
        ls.clear();
    }
    static int dfs(int tot, int root, Map<Integer, List<Integer>> mp, int par){
        int cnt = 1;
        for(int ak : mp.get(root)){
            if (ak == par) continue;
            int get = dfs(tot, ak, mp, root);
            ls.add(new int[]{get, tot - get});
            cnt += get;
        }
        return cnt;
    }

}
