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.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.StringTokenizer;

/*
1
4
1 2
2 3
3 4
11
2 2 2 2 2 3 3 7 109 109 167

DEBUG [values, [1, 2, 2]]
DEBUG [edges, [3, 3, 4]]
17

 */
public class problemD {
    static ArrayList<Edge>[] G;
    private static void solve() throws Exception {
        int n = fs.nextInt();

        Edge[] edges = new Edge[n-1];
        G = new ArrayList[n+1];
        for (int i = 1; i <= n ; i ++ ) G[i] = new ArrayList<>();

        for (int i = 0;  i < n-1 ; i ++ ) {
            int u = fs.nextInt();
            int v = fs.nextInt();
            edges[i] = new Edge(u,v);
            G[u].add(edges[i]);
            G[v].add(edges[i]);
        }
        dfs(1, 0, n);
        Arrays.sort(edges, (e1, e2) -> {
            return Long.compare(e1.w, e2.w);
        });


        int m = fs.nextInt();
        int[] p = fs.readArray(m);
        Arrays.sort(p);

        BigNumber[] values = new BigNumber[n-1];
        for (int i = 0; i < n-1; i ++ ) {
            values[i] = new BigNumber();
        }
        for (int i = 0 ; i < n-1; i ++ ) {
            values[i].multiply(i < m ? p[i] : 1);
        }
        for (int i = n-1; i < m; i ++ ) {
            values[n-2].multiply(p[i]);
        }
        Arrays.sort(values, (x1, x2) -> {
            return Double.compare(x1.log, x2.log);
        });

        long ans = 0;
        for (int i = 0; i < n-1; i ++ ) {
            ans = add(ans, mul(values[i].val, edges[i].w));
        }

        out.println(ans);
    }

    static class BigNumber {
        long val = 1;
        double log;


        public void multiply(long m) {
            val = mul(m, val);
            log = Math.log(m) + Math.log(val);
        }

        @Override
        public String toString() {
            return "{" + val + ", " + log + "}";
        }
    }

    private static long MOD = 1_000_000_000 + 7;
    private static long mul(long a, long b) {
        long c = add(a,0) * add(b,0);
        return (c >= MOD ? c % MOD : c);
    }

    private static long add(long a, long b) {
        long c = a + b;
        return (c >= MOD ? c % MOD : c);
    }

    private static long dfs(int node, int parent, int n) {
        long subTreeSize = 1;
        for (Edge e: G[node]) {
            if (e.u == parent || e.v == parent) {}
            else {
                long size = dfs(e.u + e.v - node, node, n);
                if (e.w != 0) throw null;
                e.w = size * (n-size);
                subTreeSize += size;
            }
        }
        return subTreeSize;
    }

    static class Edge{
        int u, v;
        long w;
        Edge(int a, int b) {
            u = a; v = b; w = 0;
        }

        @Override
        public String toString() {return w + "";}
    }

    private static FastScanner fs = new FastScanner();
    private static PrintWriter out = new PrintWriter(System.out);

    public static void main(String[] args) throws Exception {
        int T = 1;
        T = fs.nextInt();
        for (int t = 0; t < T; t++) {
            solve();
        }
        out.close();
    }

    static void reverse(long[] a) {
        int s = 0 ;
        int e = a.length-1;
        long temp;
        while (s <= e) {
            temp = a[s];
            a[s] = a[e];
            a[e] = temp;
            s++; e--;
        }
    }

    static void debug(Object... O) {
        System.out.print("DEBUG ");
        System.out.println(Arrays.deepToString(O));
    }

    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();
        }

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

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

        String nextString() {
            return next();
        }
    }
}
