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 D665 {
    static ArrayList<Integer> [] adj;
    static int [] sz;
    static long mod;
    public static void main(String[] args) {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        int t = sc.nextInt();
        while (t > 0) {
            int n = sc.nextInt();
            adj = new ArrayList[n + 1];
            for (int i = 1; i <= n; i++) adj[i] = new ArrayList<>();
            for (int i = 0; i < n - 1; i++) {
                int u = sc.nextInt(); int v = sc.nextInt();
                adj[u].add(v);
                adj[v].add(u);
            }
            mod = (long) 1e9 + 7;
            sz = new int[n + 1];
            dfs(1, -1);
            int m = sc.nextInt(); long num = 1;
            long [] f = new long[n + 1];
            Long [] temp = new Long[m];
            Arrays.fill(f, 1);
            for (int i = 0; i < m; i++) {
                temp[i] = sc.nextLong();
            }
            Arrays.sort(temp, Comparator.reverseOrder());
            int idx = 2;
            int total = 0;
            while (total < m - n + 1) {
                temp[m - n + 1] = (temp[m - n + 1] * temp[total]) % mod;
                total++;
            }
            for (int i = Math.max(m - n + 1, 0); i <= m - 1; i++) {
                f[idx] = temp[i]; idx++;
            }
            Integer [] a = new Integer[n];
            for (int i = 0; i < n; i++) a[i] = i + 1;
            Long [] prod = new Long[n + 1];
            for (int i = 1; i <= n; i++) prod[i] = (sz[i]) * ((long)n - sz[i]);
            Arrays.sort(prod, 2, n + 1, Comparator.reverseOrder());
            long ans = 0;
            for (int i = 2; i <= n; i++) {
                ans = (ans + prod[i] * f[i]) % mod;
            }
            out.println(ans);
            t--;
        }
        out.close();
    }

    static long pow(long b, long e) {
        long ans = 1;
        while (e > 0) {
            if (e % 2 == 1)
                ans = ans * b % mod;
            e >>= 1;
            b = b * b % mod;
        }
        return ans;
    }

    static void dfs(int cur, int par) {
        sz[cur] = 1;
        for (Integer next: adj[cur]) {
            if (next != par) {
                dfs(next, cur);
                sz[cur] += sz[next];
            }
        }
    }


    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

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

        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;
        }


    }

}