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.*;
import java.util.*;

public class D {
    //Solution by Sathvik Kuthuru
    public static void main(String[] args) {
        FastReader scan = new FastReader();
        PrintWriter out = new PrintWriter(System.out);
        Task solver = new Task();
        int t = scan.nextInt();
        for(int tt = 1; tt <= t; tt++) solver.solve(tt, scan, out);
        out.close();
    }

    static class Task {
        int n, m;
        ArrayList<Integer>[] adj;
        long[] size;
        long MOD = (long) (1e9 + 7);

        public void solve(int testNumber, FastReader scan, PrintWriter out) {
            n = scan.nextInt();
            adj = new ArrayList[n];
            size = new long[n];
            for(int i = 0; i < n; i++) adj[i] = new ArrayList<>();
            for(int i = 0; i < n - 1; i++) {
                int a = scan.nextInt() - 1, b = scan.nextInt() - 1;
                adj[a].add(b);
                adj[b].add(a);
            }
            dfs(-1, 0);
            int m = scan.nextInt();
            long[] val = new long[m];
            for(int i = 0; i < m; i++) val[i] = scan.nextLong();
            shuffle(val);
            Arrays.sort(val);
            ArrayDeque<Long> res = new ArrayDeque<>();
            for(long x : val) res.addLast(x);
            while(res.size() > n - 1) res.addLast((res.pollLast() * res.pollLast()) % MOD);
            long ans = 0;
            for(int i = 0; i < n; i++) {
                size[i] *= n - size[i];
            }
            shuffle(size);
            Arrays.sort(size);
            for(int i = n - 1; i >= 0; i--) {
                size[i] %= MOD;
                ans += size[i] * (res.isEmpty() ? 1 : res.pollLast()) % MOD;
                ans %= MOD;
            }
            out.println(ans);
        }

        void dfs(int p, int at) {
            for(int i : adj[at]) {
                if(i != p) {
                    dfs(at, i);
                    size[at] += size[i];
                }
            }
            size[at]++;
        }
    }

    static void shuffle(int[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            int temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }

    static void shuffle(long[] a) {
        Random get = new Random();
        for (int i = 0; i < a.length; i++) {
            int r = get.nextInt(a.length);
            long temp = a[i];
            a[i] = a[r];
            a[r] = temp;
        }
    }

    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(new File(s)));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

}