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.
//package maximumdistributedtree;

import java.util.*;
import java.io.*;

public class maximumdistributedtree {
	public static void main(String[] args) throws IOException {
		BufferedReader fin = new BufferedReader(new InputStreamReader(System.in));
		int t = Integer.parseInt(fin.readLine());
		StringBuilder fout = new StringBuilder();
		long mod = 1000000007;
		while(t-- > 0) {
			int n = Integer.parseInt(fin.readLine());
			ArrayList<ArrayList<Integer>> c = new ArrayList<ArrayList<Integer>>();
			long[] edgeCount = new long[n];
			for(int i = 0; i < n; i++) {
				c.add(new ArrayList<Integer>());
			}
			for(int i = 0; i < n - 1; i++) {
				StringTokenizer st = new StringTokenizer(fin.readLine());
				int a = Integer.parseInt(st.nextToken()) - 1;
				int b = Integer.parseInt(st.nextToken()) - 1;
				c.get(a).add(b);
				c.get(b).add(a);
			}
			boolean[] v = new boolean[n];
			ArrayList<Long> edges = new ArrayList<Long>();
			Queue<Integer> s = new ArrayDeque<Integer>();
			for(int i = 0; i < n; i++) {
				edgeCount[i] = c.get(i).size();
				if(edgeCount[i] == 1) {
					s.add(i);
					v[i] = true;
				}
			}
			//System.out.println(s);
			
			long[] prevNodes = new long[n];
			while(s.size() != 0) {
				int curNode = s.poll();
				long curVal = 1;
				for(int i = 0; i < c.get(curNode).size(); i++) {
					int nextNode = c.get(curNode).get(i);
					if(v[nextNode]) {
						curVal += prevNodes[nextNode];
					}
				}
				prevNodes[curNode] = curVal;
				//System.out.println("CUR NODE: " + curNode + " VAL: " + curVal);
				for(int i = 0; i < c.get(curNode).size(); i++) {
					int nextNode = c.get(curNode).get(i);
					if(prevNodes[nextNode] == 0) {
						edges.add(curVal * (n - curVal));
						edgeCount[nextNode] --;
						edgeCount[curNode] --;
					}
					if(!v[nextNode]) {
						
						if(edgeCount[nextNode] == 1) {
							v[nextNode] = true;
							s.add(nextNode);
						}	
					}
				}
			}
//			System.out.println(edges);
//			for(int i : prevNodes) {
//				System.out.print(i + " ");
//			}
//			System.out.println();
			int m = Integer.parseInt(fin.readLine());
			Long[] factors = new Long[m];
			StringTokenizer st = new StringTokenizer(fin.readLine());
			for(int i = 0; i < m; i++) {
				factors[i] = Long.parseLong(st.nextToken());
			}
			Arrays.sort(factors);
			edges.sort((a, b) -> Long.compare(a, b));
			int pointer = m - 1;
			long ans = 0;
			for(int i = edges.size() - 1; i >= 0; i--) {
				long nextFactor = 1;
				if(pointer != -1) {
					nextFactor = factors[pointer];
					pointer --;
				}
				while(pointer != -1 && pointer >= i) {
					nextFactor *= factors[pointer];
					nextFactor %= mod;
					pointer --;
				}
				ans += ((long) edges.get(i)) * nextFactor;
				ans %= mod;
			}
			fout.append(ans).append("\n");
		}
		System.out.print(fout);
	}
}
