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 Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int t=Integer.parseInt(bu.readLine());
        while(t-->0)
        {
            int n=Integer.parseInt(bu.readLine()),i;
            ArrayList<Integer> g[]=new ArrayList[n];
            for(i=0;i<n;i++)
            g[i]=new ArrayList<>();

            for(i=0;i<n-1;i++)
            {
                String s[]=bu.readLine().split(" ");
                int u=Integer.parseInt(s[0])-1,v=Integer.parseInt(s[1])-1;
                g[u].add(v);
                g[v].add(u);
            }
            int m=Integer.parseInt(bu.readLine()),a[]=new int[m];
            String s[]=bu.readLine().split(" ");
            for(i=0;i<m;i++)
            a[i]=Integer.parseInt(s[i]);
            Arrays.sort(a);

            c=new long[n];
            dfs(g,0,-1);
            c[0]=0;
            for(i=0;i<n;i++)
            c[i]=c[i]*(n-c[i]);
            Arrays.sort(c);
            long el[]=new long[n-1],M=1000000007;
            if(n-1>=m)
            {
                int j=n-2; i=m-1;
                while(i>=0)
                {
                    el[j]=a[i];
                    i--;
                    j--;
                }
                for(i=0;i<n-1;i++)
                if(el[i]==0) el[i]=1;
            }
            else
            {
                for(i=0;i<n-1;i++)
                el[i]=a[i];
                for(i=n-1;i<m;i++)
                el[n-2]=el[n-2]*a[i]%M;
            }
            long sum=0;
            i=n-1; int j=n-2;
            while(j>=0)
            {
                sum+=(c[i]*el[j])%M;
                i--;
                j--;
                sum%=M;
            }
            sb.append(sum+"\n");
        }
        System.out.print(sb);
    }

    static long c[];
    static void dfs(ArrayList<Integer> g[],int n,int p)
    {
        c[n]++;
        for(int x:g[n])
        if(x!=p)
        {
            dfs(g,x,n);
            c[n]+=c[x];
        }
    }
}