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 D {

    static ArrayList<Integer>[] tree;
    static int[] size;
    static long ans = 0;
    static int n;


    static void dfs1(int ver, int par){
        if(tree[ver].size() == 1 && tree[ver].get(0) == par){
            size[ver] = 1;
            return;
        }
        for(int z: tree[ver]){
            if(z == par) continue;
            dfs1(z, ver);
            size[ver] += size[z];
        }
        size[ver]++;
    }

    static void dfs2(int ver, int par, ArrayList<Long> contri){
        if(tree[ver].size() == 1 && tree[ver].get(0) == par) return;
        for(int z: tree[ver]){
            if(z == par) continue;
            dfs2(z, ver, contri);
            long c = (long)size[z]*(n-size[z]);
            contri.add(c);
        }
    }

    public static void main(String[] args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb = new StringBuilder();
        int tst = Integer.parseInt(br.readLine()), mod = (int)1e9+7;
        //int tst = 1;
        while(tst-->0){
            n = Integer.parseInt(br.readLine());
            tree = new ArrayList[n];
            for(int i = 0; i<n; i++) tree[i] = new ArrayList<>();
            size = new int[n];
            for(int i = 0; i<n-1; i++){
                String[] str = br.readLine().split(" ");
                int v = Integer.parseInt(str[0])-1, u = Integer.parseInt(str[1])-1;
                tree[u].add(v);
                tree[v].add(u);
            }
            ArrayList<Integer> pf = new ArrayList<>();
            int m = Integer.parseInt(br.readLine());
            StringTokenizer st = new StringTokenizer(br.readLine());
            for(int i = 0; i<m; i++){
                pf.add(Integer.parseInt(st.nextToken()));
            }
            //find size
            dfs1(0, -1);
            //find contribution of each edge
            ArrayList<Long> contri = new ArrayList<>();
            dfs2(0, -1, contri);
            long ans = 0;
            if(m>n-1){
                Collections.sort(pf);
                Collections.sort(contri);
                long p = 0;
                for(int i = 0; i<=n-3; i++){
                    p = (pf.get(i)*contri.get(i))%mod;
                    ans = (ans + p)%mod;
                }
                p = 1;
                for(int i = n-2; i<=m-1; i++){
                    p = (p*pf.get(i))%mod;
                }
                p = (p*contri.get(n-2))%mod;
                ans = (ans + p)%mod;
            }
            else{
                Collections.sort(pf, Collections.reverseOrder());
                Collections.sort(contri, Collections.reverseOrder());
                long p = 0;
                for(int i = 0; i<m; i++){
                    p = (pf.get(i)*contri.get(i))%mod;
                    ans = (ans + p)%mod;
                }
                p = 1;
                for(int i = m; i<n-1; i++){
                    ans = (ans + contri.get(i))%mod;
                }
            }
            sb.append(ans).append('\n');
        }
        System.out.println(sb);
    }
}
