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 Main
{
    static ArrayList<Integer> edge[];
    static ArrayList<Long> freq;
    public static long dfs(int u,boolean vis[],int n)
    {
        long ans=1;
        vis[u]=true;
        for(int i:edge[u])
        {
            if(!vis[i])
            {
                long nn=dfs(i,vis,n);
                freq.add(nn*(n-nn));
                ans+=nn;
            }
        }
        return ans;
    }
    public static void main(String arfs[])throws Exception
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw=new PrintWriter(System.out);
        final long mod=(long)1e9+7;
        int t=Integer.parseInt(br.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(br.readLine());
            edge=new ArrayList[n+1];
            for(int i=1;i<=n;i++)
            edge[i]=new ArrayList<>();
            for(int i=1;i<n;i++)
            {
                String str[]=br.readLine().split(" ");
                int u=Integer.parseInt(str[0]);
                int v=Integer.parseInt(str[1]);
                edge[u].add(v);
                edge[v].add(u);
            }
            int m=Integer.parseInt(br.readLine());
            long factor[]=new long[m];
            String str[]=br.readLine().split(" ");
            for(int i=0;i<m;i++)
            factor[i]=Long.parseLong(str[i]);
            freq=new ArrayList<>();
            dfs(1,new boolean[n+1],n);
            Collections.sort(freq);
            Arrays.sort(factor);
            long ans=0;
            if(m<=n-1)
            {
                int j=n-2;
                for(int i=m-1;i>=0;i--,j--)
                ans=(ans+(freq.get(j)*factor[i])%mod)%mod;
                for(;j>=0;j--)
                ans=(ans+(freq.get(j)*1L)%mod)%mod;
            }
            else
            {
                for(int i=m-2;i>=n-2;i--)
                factor[i]=(factor[i]*factor[i+1])%mod;
                for(int i=n-2;i>=0;i--)
                ans=(ans+(freq.get(i)*factor[i])%mod)%mod;
            }
            pw.println(ans);
        }
        pw.flush();
        pw.close();
    }
}