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.io.*;

public class Question4My {
    
	static HashSet<Integer> nodes[];
	static boolean vis[];
	static int n;
	static int subtreeSize[];
	static ArrayList<Long> list;
	static int MOD = (int)(1e9 + 7);
	static Reader sc = new Reader();
	public static void main(String[] args) throws IOException{
		int t = sc.nextInt();
		while(t-->0) {
			solve();
		}
	}
	
	public static void solve() throws IOException {
		
	    n = sc.nextInt();
		nodes = new HashSet[n];
		vis = new boolean[n];
		subtreeSize = new int[n];
		list = new ArrayList<>();
		
		//initiate
		for(int i = 0;i < n;i++)
			nodes[i] = new HashSet<Integer>();
		
		for(int i = 0;i < n - 1;i++) {
			int u = sc.nextInt() - 1;
			int v = sc.nextInt() - 1;
			
			nodes[u].add(v);
			nodes[v].add(u);
		}
	
		//input prime factors
		int m = sc.nextInt();
		ArrayList<Integer> factors = new ArrayList<>();
		
		for(int i = 0;i < m;i++) {
			factors.add(sc.nextInt());
		}
		
		while(m < n - 1) {
			factors.add(1);
			m++;
		}
		

		Collections.sort(factors);
		dfs(0);
		Collections.sort(list);
//		System.out.println(list);
//		System.out.println(factors);
//		System.out.println("\n");
//		
//		System.out.println("list size : " + list.size());
//		System.out.println("factors size : " + factors.size());
		int a = list.size() - 1;
		long ans = 0;
		long[] values = new long[list.size()];
		values[a] = (long)list.get(a);
		for(int i = a + 1;i < factors.size();i++) {
			values[a] = (values[a] * factors.get(i)) % MOD;
		}
		
		for(int i = 0;i < a;i++)values[i] = (long)list.get(i);
		
		for(int i = 0;i < values.length;i++) {
			ans = (ans + (values[i] * factors.get(i)) % MOD) % MOD;
		}
		System.out.println(ans);
		
	}
	
	public static void dfs(int curr) {
		vis[curr] = true;
		
		int subtree = 1;
		for(Integer ix : nodes[curr]) {
			if(!vis[ix]) {
				dfs(ix);
				subtree += subtreeSize[ix];
			}
		}
		
		subtreeSize[curr] = subtree;
		
		if((long)subtree * (long)(n - subtree) != 0)
		list.add((long)subtree * (long)(n - subtree));
		
		
	}
	
	static long mul(long a, long b) {
		return (a * b) % MOD;
	}
	
		
	
	static class Reader 
    { 
        final private int BUFFER_SIZE = 1 << 16; 
        private DataInputStream din; 
        private byte[] buffer; 
        private int bufferPointer, bytesRead; 
  
        public Reader() 
        { 
            din = new DataInputStream(System.in); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public Reader(String file_name) throws IOException 
        { 
            din = new DataInputStream(new FileInputStream(file_name)); 
            buffer = new byte[BUFFER_SIZE]; 
            bufferPointer = bytesRead = 0; 
        } 
  
        public String readLine() throws IOException 
        { 
            byte[] buf = new byte[64]; // line length 
            int cnt = 0, c; 
            while ((c = read()) != -1) 
            { 
                if (c == '\n') 
                    break; 
                buf[cnt++] = (byte) c; 
            } 
            return new String(buf, 0, cnt); 
        } 
  
        public int nextInt() throws IOException 
        { 
            int ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do
            { 
                ret = ret * 10 + c - '0'; 
            }  while ((c = read()) >= '0' && c <= '9'); 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public long nextLong() throws IOException 
        { 
            long ret = 0; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        public double nextDouble() throws IOException 
        { 
            double ret = 0, div = 1; 
            byte c = read(); 
            while (c <= ' ') 
                c = read(); 
            boolean neg = (c == '-'); 
            if (neg) 
                c = read(); 
  
            do { 
                ret = ret * 10 + c - '0'; 
            } 
            while ((c = read()) >= '0' && c <= '9'); 
  
            if (c == '.') 
            { 
                while ((c = read()) >= '0' && c <= '9') 
                { 
                    ret += (c - '0') / (div *= 10); 
                } 
            } 
  
            if (neg) 
                return -ret; 
            return ret; 
        } 
  
        private void fillBuffer() throws IOException 
        { 
            bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE); 
            if (bytesRead == -1) 
                buffer[0] = -1; 
        } 
  
        private byte read() throws IOException 
        { 
            if (bufferPointer == bytesRead) 
                fillBuffer(); 
            return buffer[bufferPointer++]; 
        } 
  
        public void close() throws IOException 
        { 
            if (din == null) 
                return; 
            din.close(); 
        } 
    } 
	
}
