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.Comparator;
import java.util.LinkedList;
import java.util.StringTokenizer;


public class Learning {
    static ArrayList<Long> p;
    static int n;

    public static void main(String[] args) throws Exception {
        StringBuilder st = new StringBuilder();
        FastInput in = new FastInput();
        int t = in.nextInt();
        while (t-- > 0) {
            n = in.nextInt();
            LinkedList<Integer>[] adj = new LinkedList[n];
            for (int i = 0; i < n; i++) {
                adj[i] = new LinkedList<>();
            }
            for (int i = 0; i < n - 1; i++) {
                int c = in.nextInt() - 1;
                int u = in.nextInt() - 1;
                adj[c].add(u);
                adj[u].add(c);
            }
            int m = in.nextInt();
            ArrayList<Long> b = new ArrayList<>();
            p = new ArrayList<>();
            for (int i = 0; i < m; i++) {
                b.add(in.nextLong());
            }
            dfs(0, -1, adj);
            b.sort(Comparator.reverseOrder());
            p.sort(Comparator.reverseOrder());
            long x = 1;
            int mod = 1000000007;
            int i, j = 0;
            for (i = 0; i <= m - (n - 1); i++) {
                x = (x * b.get(i)) % mod;
            }
            if (i > 0) {
                i--;
                b.set(i, x);
            }
            long ans = 0;
            while (i < m && j < (n - 1)) {
                ans += (b.get(i++) * p.get(j++)) % mod;
                ans %= mod;
            }
            while (j < n - 1) {
                ans += p.get(j++);
                ans %= mod;
            }
            st.append(ans).append("\n");

        }
        System.out.print(st);

    }

    private static long dfs(int i, int par, LinkedList<Integer>[] adj) {
        long ans = 1;
        for (int v : adj[i]) {
            if (v != par)
                ans += dfs(v, i, adj);
        }

        p.add(ans * (n - ans));

        return ans;
    }

    private static int gcd(int a, int b) {
        int max = Math.max(a, b);
        int min = Math.min(a, b);
        if (min == 0) {
            return max;
        }
        return gcd(max % min, min);
    }

}


class FastInput {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st;

    String next() throws IOException {
        if (st == null || !st.hasMoreTokens())
            st = new StringTokenizer(br.readLine());
        return st.nextToken();
    }

    Integer nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    Long nextLong() throws IOException {
        return Long.parseLong(next());
    }
}