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 CFD {
    BufferedReader br;
    PrintWriter out;
    StringTokenizer st;
    boolean eof;
    private static final long MOD = 1000L * 1000L * 1000L + 7;
    private static final int[] dx = {0, -1, 0, 1};
    private static final int[] dy = {1, 0, -1, 0};
    private static final String yes = "Yes";
    private static final String no = "No";

    void solve() {
        int T = nextInt();
        for (int i = 0; i < T; i++) {
            helper();
        }
    }

    List<List<Integer>> graph;
    int n;
    int[] subtree;
    List<Long> edgeCnt;
    int LIMIT = 100 * 1000;
    int[] primeCnt;
    void helper() {
        n = nextInt();
        graph = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            graph.add(new ArrayList<>());
        }
        for (int i = 0; i < n - 1; i++) {
            int from = nextInt() - 1;
            int to = nextInt() - 1;
            graph.get(from).add(to);
            graph.get(to).add(from);
        }
        subtree = new int[n];
        dfs(-1, 0);

        edgeCnt = new ArrayList<>();
        dfs2(0, 0);

        Collections.sort(edgeCnt);
        Collections.reverse(edgeCnt);

        int m = nextInt();

        primeCnt = new int[LIMIT];
        for (int i = 0; i < m; i++) {
            int p = nextInt();
            factor(p);
        }

        List<Integer> allPrime = new ArrayList<>();
        for (int i = 0; i < LIMIT; i++) {
            for (int j = 0; j < primeCnt[i]; j++) {
                allPrime.add(i);
            }
        }

        Collections.reverse(allPrime);
        long[] multi = new long[n - 1];
        Arrays.fill(multi, 1);

        if (n - 1 >= allPrime.size()) {
            int sz = allPrime.size();
            for (int i = 0; i < sz; i++) {
                multi[i] = allPrime.get(i);
            }
        } else {
            int left = allPrime.size() - (n - 1);
            long tmp = 1;
            for (int i = 0; i < left; i++) {
                tmp = tmp * allPrime.get(i);
                tmp = tmp % MOD;
            }
            multi[0] = tmp;
            for (int i = 0; i < n - 1; i++) {
                multi[i] = multi[i] * allPrime.get(left + i);
                multi[i] = multi[i] % MOD;
            }
        }

        long res = 0;
        for (int i = 0; i < n - 1; i++) {
            res += multi[i] * edgeCnt.get(i) % MOD;
            if (res >= MOD) {
                res -= MOD;
            }
        }

        outln(res % MOD);
    }

    // factorization of a long num
    public List<Integer> factor(int N) {
        List<Integer> res = new ArrayList<>();
        for (int i = 2; i * i <= N; i++) {
            while (N % i == 0) {
                res.add(i);
                primeCnt[i]++;
                N = N / i;
            }
        }
        if (N != 1L) {
            res.add(N);
            primeCnt[N]++;
        }

        return res;
    }

    void dfs2(int par, int cur) {
        if (par != cur) {
            edgeCnt.add(1L * subtree[cur] * (n - subtree[cur]));
        }
        for (int nxt : graph.get(cur)) {
            if (nxt == par) {
                continue;
            }
            dfs2(cur, nxt);
        }
    }

    void dfs(int par, int cur) {
        subtree[cur]++;
        for (int nxt : graph.get(cur)) {
            if (nxt == par) {
                continue;
            }
            dfs(cur, nxt);
            subtree[cur] += subtree[nxt];
        }
    }

    void shuffle(long[] a) {
        int n = a.length;
        for(int i = 0; i < n; i++) {
            int r = i + (int) (Math.random() * (n - i));
            long tmp = a[i];
            a[i] = a[r];
            a[r] = tmp;
        }
    }
    long gcd(long a, long b) {
        while(a != 0 && b != 0) {
            long c = b;
            b = a % b;
            a = c;
        }
        return a + b;
    }
    private void outln(Object o) {
        out.println(o);
    }
    private void out(Object o) {
        out.print(o);
    }
    private void formatPrint(double val) {
        outln(String.format("%.9f%n", val));
    }
    public CFD() {
        br = new BufferedReader(new InputStreamReader(System.in));
        out = new PrintWriter(System.out);
        solve();
        out.close();
    }
    public static void main(String[] args) {
        new CFD();
    }

    public long[] nextLongArr(int n) {
        long[] res = new long[n];
        for(int i = 0; i < n; i++)
            res[i] = nextLong();
        return res;
    }
    public int[] nextIntArr(int n) {
        int[] res = new int[n];
        for(int i = 0; i < n; i++)
            res[i] = nextInt();
        return res;
    }
    public String nextToken() {
        while (st == null || !st.hasMoreTokens()) {
            try {
                st = new StringTokenizer(br.readLine());
            } catch (Exception e) {
                eof = true;
                return null;
            }
        }
        return st.nextToken();
    }
    public String nextString() {
        try {
            return br.readLine();
        } catch (IOException e) {
            eof = true;
            return null;
        }
    }
    public int nextInt() {
        return Integer.parseInt(nextToken());
    }
    public long nextLong() {
        return Long.parseLong(nextToken());
    }
    public double nextDouble() {
        return Double.parseDouble(nextToken());
    }
}