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 Main {
    public static PrintWriter out;
    public static ArrayList <Integer>[] adjacencyList;
    public static ArrayList<Integer> p;
    public static ArrayList<Long> sizeList;
    public static long mod = (long)1e9+7;
    public static void main(String[] args){
        MyScanner sc = new MyScanner();
        out = new PrintWriter(new BufferedOutputStream(System.out));
        /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();     // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */
        int t=sc.nextInt();
        while(t-- >0){
            int n=sc.nextInt();
            adjacencyList=new ArrayList[n];
            for(int i=0;i<n;i++)adjacencyList[i]=new ArrayList<>();
            for(int i=0;i<n-1;i++){
                int u,v;u=sc.nextInt();v=sc.nextInt();u--;v--;
                adjacencyList[u].add(v);adjacencyList[v].add(u);
            }
            int m=sc.nextInt();
            p=new ArrayList<>();
            for(int i=0;i<m;i++)p.add(sc.nextInt());
            sizeList=new ArrayList<>();
            dfs(0,0,n);
            Collections.sort(sizeList,Collections.reverseOrder());
            Collections.sort(p,Collections.reverseOrder());
            long tmp=1l;
            int indx=0;
            long ans=0;
            if(m>n-1){
                for(int i=0;i<m-n+1;i++){tmp*=p.get(i);tmp%=mod;}
                indx=m-n+1;
               // ans+=tmp*
            }
            for(int i=indx;i<m;i++){
               // if(i==indx)
                ans+=1l*sizeList.get(i-indx)*p.get(i);
                ans%=mod;
                //else
                if(i==indx)ans*=tmp;
                ans%=mod;
            }
            for(int i=m;i<n;i++){ans+=1l*sizeList.get(i);ans%=mod;}
           // ans+=
            out.println(ans);
        }
        out.close();
    }
    public static int dfs(int root,int prnt,int n){
      int sz=1;
      for(int i=0;i<adjacencyList[root].size();i++){
          if(prnt==adjacencyList[root].get(i))continue;
          sz+=dfs(adjacencyList[root].get(i),root,n);
      }
      sizeList.add(1l*sz*(n-sz));
      return  sz;
    }
    //-----------MyScanner class for faster input----------
    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }

        long nextLong() {
            return Long.parseLong(next());
        }

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

    }
}
