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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;

public class d1 extends PrintWriter {

        static BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
        // static Scanner s=new Scanner(System.in);
        d1() { super(System.out); }
        public static void main(String[] args) throws IOException{
            d1 d1=new d1();d1.main();d1.flush();

        }
        void main() throws IOException {
            BufferedReader s = new BufferedReader(new InputStreamReader(System.in));
            StringBuffer sb = new StringBuffer();
            StringBuffer sb1 = new StringBuffer();
            PrintWriter out = new PrintWriter(System.out);
        //    Scanner s1=new Scanner(System.in);
            String[] s1=s();
            int t=i(s1[0]);
            while(t-->0){
                //for(int i=0;i<n;i++)
                //int[] a=new int[n];
                String[] s2=s();
                n=i(s2[0]);
                ArrayList<Integer> []adj=new ArrayList[n+1];
                for(int i=0;i<n-1;i++){
                    String[] s3=s();
                   int u=i(s3[0]);int v=i(s3[1]);
                   if(adj[u]==null) adj[u]=new ArrayList<>();
                   if(adj[v]==null) adj[v]=new ArrayList<>();
                   adj[u].add(v);adj[v].add(u);
                }
                int m=i(s()[0]);
                String[] s4=s();
                long[] p=new long[m];for(int i=0;i<m;i++) p[i]=i(s4[i]);
               arr=new ArrayList<>();node=new long[n+1];
                dfs(1,new int[n+1],adj);
                long[] val=new long[arr.size()];int j=0;
               for(long i1:arr) val[j++]=i1;
               Arrays.sort(val);
               j--;long ans=0;int mod=1000000007;
                if(m>n-1){
                    Arrays.sort(p);
                    long tm=1;
                    for(int i=m-1;i>=0;i--){
                        if(i+1>n-1) tm=(tm*p[i])%mod;
                        else{
                            tm=(tm*p[i])%mod;
                            ans=(ans%mod+(tm*(val[j--]%mod)))%mod;
                            tm=1;
                        }
                    }
                }else{
                    Arrays.sort(p);
                    for(int i=m-1;i>=0;i--){
                      //  System.out.println(val[j]);
                            ans=(ans%mod+((p[i]%mod)*(val[j--]%mod)))%mod;
                    }
                    while(j>=0){
              //          System.out.println(val[j]);
                        ans=(ans%mod+val[j--]%mod)%mod;
                    }
                }
                System.out.println(ans);
            }
           }static ArrayList<Long> arr;static int n;static long[] node;
           static long dfs(int i,int[] vis,ArrayList<Integer>[] adj){
            vis[i]=1;long nodes=0;
            for(Integer j:adj[i]){
                if(vis[j]==0){
                    nodes+=dfs(j,vis,adj);
                }else nodes+=node[j];
            }node[i]=nodes+1;
            //   System.out.println("HI"+" "+nodes+" "+n);
       arr.add(node[i]*(n-node[i]));
            return nodes+1;
           }
    static long gcd(long a, long b) {
        if (b == 0)
            return a;
        return gcd(b, a % b);
    }
        static String[] s() throws IOException {
            return s.readLine().trim().split("\\s+");
        }

        static int i(String ss) {
            return Integer.parseInt(ss);
        }
        static long l(String ss) {
            return Long.parseLong(ss);
        }
    }



