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.File;
import java.io.FileNotFoundException;
import java.util.*;

public class D {
    static int mod = 1000000007;
    public static void main(String[] args) {
        Scanner input;
        try {
            input = new Scanner(new File("input.txt"));
        } catch (FileNotFoundException e) {
            input = new Scanner(System.in);
        }

        int Tests = input.nextInt();
        while (Tests-- > 0) {
            int n = input.nextInt();
            ArrayList<Integer> edges[] = new ArrayList[n+1];
            for (int i = 1; i <= n; i++) edges[i] = new ArrayList();
            for (int i = 1; i < n; i++) {
                int u = input.nextInt();
                int v = input.nextInt();
                edges[u].add(v);
                edges[v].add(u);
            }
            int m = input.nextInt();
            ArrayList<Integer> factors = new ArrayList();
            for (int i = 1; i <= m; i++) {
                factors.add(input.nextInt());
            }
            while (factors.size() < n-1) {
                factors.add(1);
            }
            
            ArrayList<Long> coef = new ArrayList();
            int[] sz = new int[n+1];
            DFS(1, -1, n, edges, coef, sz);


            Collections.sort(factors);
            Collections.sort(coef);
            long ans = 0;
            for (int i = 0; i < n-2; i++) {
                ans += (long)factors.get(i)*(coef.get(i)%mod);
                ans %= mod;
            }
            long factor_end = 1;
            for (int i = n-2; i < factors.size(); i++) {
                factor_end *= factors.get(i);
                factor_end %= mod;
            }
            ans += (long)factor_end*(coef.get(n-2) % mod);
            ans %= mod;
            System.out.println(ans);
        }
        input.close();
    }
    static void DFS(int u, int p, int n, ArrayList<Integer> g[], ArrayList<Long> coef, int sz[]) {
        sz[u] = 1;
        for (int v : g[u]) {
            if (v == p) continue;
            DFS(v, u, n, g, coef, sz);
            sz[u] += sz[v];
        }
        if (u != 1) {
            coef.add((long)sz[u]*(n - sz[u]));
        }
    }
}
