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.lang.*;
import java.io.*;

public class Main
{
	PrintWriter out = new PrintWriter(System.out);
	BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer tok = new StringTokenizer("");
    String next() throws IOException {
        if (!tok.hasMoreTokens()) { tok = new StringTokenizer(in.readLine()); }
        return tok.nextToken();
    }
    int ni() throws IOException { return Integer.parseInt(next()); }
    long nl() throws IOException { return Long.parseLong(next()); }
    
    int n,root,maxdepth;
    long mod=1000000007;
    ArrayList<Integer>[]A;
    long[]P,E,S;
    
    void solve() throws IOException {
        for (int tc=ni();tc>0;tc--) {
            n=ni();
            A=new ArrayList[n+1];
            for (int i=1;i<=n;i++) A[i]=new ArrayList();
            for (int i=1;i<n;i++) {
                int u=ni(),v=ni();
                A[u].add(v);
                A[v].add(u);
            }
            
            root=1;
            maxdepth=0;
            dfs1(1,0,1);
            
            S=new long[n+1];
            E=new long[n+1];
            P=new long[n+1];
            int m=ni();
            Arrays.fill(P,1);
            P[0]=0;
            P[1]=0;
            if (m<=n-1) {
                for (int i=2;i<=m+1;i++) P[i]=nl();
                Arrays.sort(P);
            }
            else {
                long[] P2=new long[m];
                for (int i=0;i<m;i++) P2[i]=nl();
                Arrays.sort(P2);
                int j=2;
                P[n]=1;
                for (int i=0;i<m;i++) {
                    if (j<n) P[j]=P2[i];
                    else P[n]=(P[n]*P2[i])%mod;
                    j++;
                }
                
            }
            dfs2(root,0);
            
            Arrays.sort(E);
            //Arrays.sort(P);
            long ans=0;
            for (int i=2;i<=n;i++) ans=(ans+P[i]*(E[i]%mod))%mod;
            out.println(ans);
        }
        out.flush();
    }
    
    void dfs2(int u,int p) {
        S[u]=1;
        for (Integer v:A[u]) {
            if (v==p) continue;
            dfs2(v,u);
            S[u]+=S[v];
        }
        E[u]=(S[u]*(n-S[u]));
    }
    
    void dfs1(int u,int p,int d) {
        if (d>maxdepth) { maxdepth=d; root=u; }
        for (Integer v:A[u]) {
            if (v==p) continue;
            dfs1(v,u,d+1);
        }
    }
    
    public static void main(String[] args) throws IOException {
        new Main().solve();
    }
}