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.lang.Math;
import java.util.*;

public class Main  {

    public BufferedReader in;
    public PrintStream out;

    public boolean log_enabled = false;
    
    public boolean multiply_tests = true;

    private class TestCase {

        int n;
        int[] childs_count;
        int[][] childs;

        long[] esum;
        int ek;

        int[][] adj;

        public long dfs(int u, int r)
        {
            long  s = 0, k;
            int i;
            
            for (i=0; i<childs_count[u]; i++)
            {
                if (childs[u][i] != r)
                {
                    k = dfs(childs[u][i], u);
                    esum[ek++] = k * (n-k);
                    s += k;
                }
            }

            return s+1;
        }

        public Object solve() {
            
            n = readInt();

            childs_count = new int[n];
            childs = new int[n][];
            
            adj = new int[n-1][2];
            
            Arrays.fill(childs_count, 0);

            int i;
            for (i=0; i<n-1; i++)
            {
                adj[i][0] = readInt()-1;
                adj[i][1] = readInt()-1;
                
                childs_count[adj[i][0]] ++;
                childs_count[adj[i][1]] ++;
            }
            
            for (i=0; i<n; i++)
            {
                childs[i] = new int[childs_count[i]];
                childs_count[i] = 0;
            }
            
            int u,v;
            for (i=0; i<n-1; i++)
            {
                u = adj[i][0];
                v = adj[i][1];
                
                childs[u][ childs_count[u] ++ ] = v;
                childs[v][ childs_count[v] ++ ] = u;
            }

            esum = new long[n-1];
            ek = 0;
            
            dfs(0, -1);

            Arrays.sort(esum);
            
            int m = readInt();
            long[] p = readLongArray(m);

            Arrays.sort(p);

            long M = 1000000007;
            long R = 0;

            if (m > n-1)
            {
                for (i=n-1; i<m; i++)
                {
                    p[n-2] = (p[n-2] * p[i]) % M;
                }

                m = n-1;
            }
                          
            for (i=0; i<m; i++)
            {
                R += (p[m-i-1] * esum[n-i-2]) % M;
                R = R % M;
            }

            for (i=m; i<n-1; i++)
            {
                R = (R + esum[n-i-2]) % M;
            }
            
            return R;
            
            //return strf("%f", 0);
            
            //out.printf("Case #%d: \n", caseNumber);
            //return null;
        }
        
        public int caseNumber;
        
        TestCase(int number) {
            caseNumber = number;
        }
        
        public void run(){
            Object r = this.solve();
            
            if ((r != null))
            {
                //outputCaseNumber(r);
                out.println(r);
            }
        }
        
        public String impossible(){
            return "IMPOSSIBLE";
        }
        
        public String strf(String format, Object... args)
        {
            return String.format(format, args);
        }
        
//        public void outputCaseNumber(Object r){
//            //out.printf("Case #%d:", caseNumber);
//            if (r != null)
//            {
//              //  out.print(" ");
//                out.print(r);
//            }
//            out.print("\n");
//        }
    }

    public void run() {
        //while (true)
        {
            int t = multiply_tests ?  readInt() : 1;
            for (int i = 0; i < t; i++) {
                TestCase T = new TestCase(i + 1);
                T.run();
            }
        }
    }
    

    
    public Main(BufferedReader _in, PrintStream _out){
        in = _in;
        out = _out;
    }
    

    public static void main(String args[]) {
        Locale.setDefault(Locale.US);
        Main S;
        try {
            S = new Main(
                        new BufferedReader(new InputStreamReader(System.in)),
                        System.out
                );
        } catch (Exception e) {
            return;
        }
        
        S.run();
        
    }

    private StringTokenizer tokenizer = null;

    public int readInt() {
        return Integer.parseInt(readToken());
    }

    public long readLong() {
        return Long.parseLong(readToken());
    }

    public double readDouble() {
        return Double.parseDouble(readToken());
    }

    public String readLn() {
        try {
            String s;
            while ((s = in.readLine()).length() == 0);
            return s;
        } catch (Exception e) {
            return "";
        }
    }

    public String readToken() {
        try {
            while (tokenizer == null || !tokenizer.hasMoreTokens()) {
                tokenizer = new StringTokenizer(in.readLine());
            }
            return tokenizer.nextToken();
        } catch (Exception e) {
            return "";
        }
    }

    public int[] readIntArray(int n) {
        int[] x = new int[n];
        readIntArray(x, n);
        return x;
    }

    public void readIntArray(int[] x, int n) {
        for (int i = 0; i < n; i++) {
            x[i] = readInt();
        }
    }
    
    public long[] readLongArray(int n) {
        long[] x = new long[n];
        readLongArray(x, n);
        return x;
    }

    public void readLongArray(long[] x, int n) {
        for (int i = 0; i < n; i++) {
            x[i] = readLong();
        }
    }

    public void logWrite(String format, Object... args) {
        if (!log_enabled) {
            return;
        }

        out.printf(format, args);
    }
}
