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.util.ArrayList;
import java.util.Arrays;
import java.util.Collection;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.StringTokenizer;

public class D {
	
	static int MOD = 1_000_000_007;
	static HashMap<Integer, ArrayList<Integer>> map;
	static HashMap<String, Long> nm;
	static ArrayList<Long> nl;
	static Long vis[], num[];
	static long dp[][];

	public static void main(String args[]) throws NumberFormatException, IOException {
		
		BufferedReader bi = new BufferedReader(new InputStreamReader(System.in));
		int n = Integer.parseInt(bi.readLine());
		int l, m, u, v;
		String str;
		ArrayList<Long> facts;
		long ans;
		
		for(int i = 0; i < n; i++) {
			
			l = Integer.parseInt(bi.readLine());
			map = new HashMap<Integer, ArrayList<Integer>>();
			facts = new ArrayList<Long>();
			nm = new HashMap<String, Long>();
			num = new Long[2 * l + 1];
			vis = new Long[l+1];
			nl = new ArrayList<Long>();
			
			for(int j = 0; j < l-1; j++) {
				
				str = bi.readLine();
				
				StringTokenizer st = new StringTokenizer(str, " ");		
				
				u = Integer.parseInt(st.nextToken());
				v = Integer.parseInt(st.nextToken());
				
				if(!map.containsKey(u)) {
					map.put(u, new ArrayList<Integer>());					
				}
				map.get(u).add(v); 
				if(!map.containsKey(v)) {
					map.put(v, new ArrayList<Integer>());					
				}
				map.get(v).add(u); 
				
			}
			
			m = Integer.parseInt(bi.readLine());
				
			str = bi.readLine();
				
			StringTokenizer st = new StringTokenizer(str, " ");
				
			while (st.hasMoreTokens()) {
				
				facts.add(Long.parseLong(st.nextToken()));
				    
			} 
			
			Collections.sort(facts);
			
			while(facts.size() > l-1) {
				
				long tmp = facts.remove(facts.size()-1);
				facts.set(facts.size() - 1, (facts.get(facts.size()-1) * tmp) % MOD);
				
			}		
				
			Arrays.fill(vis, 0L);
			Arrays.fill(num, 0L);
			
			dfs(1);
				
			ans = 0L;
			
			for(int j = 0; j < nl.size(); j++) {
				
				nl.set(j, (l - nl.get(j)) * nl.get(j));
				
			}
						
			Collections.sort(nl, Collections.reverseOrder());
			
			for(long ln : nl) {				
								
				if(facts.size() > 0) {
					ans = (ans + (facts.get(facts.size()-1) * (ln % MOD)) % MOD) % MOD;
					facts.remove(facts.size()-1);
				}else {
					ans = (ans + (ln % MOD)) % MOD;					
				}
				
			}
			
			System.out.println(ans);			
			
		}
		
		
	}
	
	static long dfs(int s) {
		
		vis[s] = 1L;
		
		long ret = 1;
								
		ArrayList<Integer> tmp = map.get(s);	
						
		for(int i = 0; i < tmp.size(); i++) {
			
			if(vis[tmp.get(i)] != 1) {
				
				nl.add(dfs(tmp.get(i)));
				ret += nl.get(nl.size()-1);
				
			}
			
		}
		
		return ret;
		
	}
	
	/*static long dfs(int s, int f, int m) {
				
		ArrayList<Integer> tmp = map.get(s);
		
		long ret = 0;
		
		if(s > m)
			ret = 1;
		
		for(int i = 0; i < tmp.size(); i++) {
			
			long tm = 0;
			
			if(tmp.get(i) != f && (tm = dfs(tmp.get(i), s, m)) > 0) {
								
				//System.out.println(tmp.get(i) + " " + s);
				
				String tmpkey = "" + Math.min(s, tmp.get(i)) + "" + Math.max(s, tmp.get(i));
				
				if(nm.containsKey(tmpkey)) {
					
					nm.replace(tmpkey, (nm.get(tmpkey)+tm) % MOD);
					
				}else {
					
					nm.put(tmpkey, tm);
					
				}
				
				ret += tm;
				
			}
			
		}	
		
		return ret;
		
	}*/
	
}