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 Main {
    static int M = 1_000_000_007;

    private static int testCase(int n, int[][] edges, int m, int[] p) {
        ArrayList<Integer>[] adjList = new ArrayList[n];
        ArrayList<Long> pComp = new ArrayList<>();
        boolean[] visited = new boolean[n];
        PriorityQueue<Long> weights = new PriorityQueue<>((a, b) -> Long.compare(b, a));
        long ans = 0;

        for (int pi : p) pComp.add((long) pi);

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

        for (int[] edge : edges) {
            adjList[edge[0] - 1].add(edge[1] - 1);
            adjList[edge[1] - 1].add(edge[0] - 1);
        }

        dfs(adjList, 0, visited, weights, n);

        Collections.sort(pComp);

        while (pComp.size() > n - 1) {
            long next = pComp.get(pComp.size() - 1) * pComp.get(pComp.size() - 2);
            pComp.remove(pComp.size() - 1);
            pComp.remove(pComp.size() - 1);
            pComp.add(next % M);
        }

        assert weights.size() == n - 1;
        assert pComp.size() <= weights.size();

        int idx = pComp.size() - 1;

        while (!weights.isEmpty()) {
            ans = (ans + ((weights.poll() % M) * (idx == -1 ? 1 : pComp.get(idx--)))) % M;
        }

        return (int) ans;
    }

    private static int dfs(ArrayList<Integer>[] adjList, int curr, boolean[] visited, PriorityQueue<Long> weights, int n) {
        visited[curr] = true;
        int numChildrenTotal = 1, numChildren;

        for (int v : adjList[curr]) {
            if (!visited[v]) {
                numChildren = dfs(adjList, v, visited, weights, n);

                weights.offer((long) numChildren * (n - numChildren));
                numChildrenTotal += numChildren;
            }
        }

        return numChildrenTotal;
    }

    public static void main(String[] args) {
        FastScanner in = new FastScanner();
        PrintWriter out = new PrintWriter(System.out);
        int t = in.nextInt();  // Scanner has functions to read ints, longs, strings, chars, etc.
        //in.nextLine();
        for (int tt = 1; tt <= t; ++tt) {
            int n = in.nextInt();
            int[][] edges = new int[n - 1][2];

            for (int i = 0; i < n - 1; i++) {
                edges[i] = in.readArray(2);
            }

            int m = in.nextInt();
            int[] p = in.readArray(m);

            out.println(testCase(n, edges, m, p));
        }

        out.close();
    }

    private static class FastScanner {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        char[] readCharArray(int n) {
            char[] arr = new char[n];
            try {
                br.read(arr);
                br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return arr;
        }

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

        int[] readArray(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) a[i] = nextInt();
            return a;
        }

        long[] readLongArray(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) a[i] = nextLong();
            return a;
        }

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