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.ArrayList;
import java.util.Arrays;
import java.util.Scanner;

public class MaxSum_All_NodeToNode_Dist {
    static ArrayList<Integer>[] g;
    static long[] subTree;
    static long MOD = 1000000007;
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int T = scan.nextInt();
        while (T-- > 0) {
            int n = scan.nextInt();
            init(n);
            for (int i = 0; i < n-1; i++) addEdge(scan.nextInt(), scan.nextInt());
            dfs(1, 0);
            long[] usage = new long[n-1]; int ind = 0;
            for (int i = 2; i <= n; i++) usage[ind++] = subTree[i] * (n - subTree[i]);
            Arrays.sort(usage);

            int m = scan.nextInt();
            ArrayList<Long> primes = new ArrayList<>();
            for (int i = 0; i < m; i++) primes.add(scan.nextLong());
            for (int i = m; i < n-1; i++) primes.add((long) 1);
            primes.sort((a,b) -> (int) (a-b));
            for (int i = m; i >  n-1; i--) {
                long p1 = primes.get(primes.size() - 1);
                primes.remove(primes.size() - 1);
                long p2 = primes.get(primes.size() - 1);
                primes.remove(primes.size() - 1);
                primes.add(p1 * p2 % MOD);
            }
            long ans = 0;
            for (int i = 0; i < n-1; i++) {
                ans += (usage[i] * primes.get(i)) % MOD;
                ans = ans % MOD;
            }
            System.out.println(ans);
        }
    }
    static void dfs(int u, int p) {
        subTree[u] = 1;
        for (int v: g[u]) {
            if (v != p) {
                dfs(v, u);
                subTree[u] += subTree[v];
            }
        }
    }
    static void addEdge (int u, int v) {
        g[u].add(v); g[v].add(u);
    }
    static void init(int n) {
        g = new ArrayList[n+1];
        for (int i = 0; i <= n; i++) g[i] = new ArrayList<>();
        subTree = new long[n+1];
    }
}
