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 sys,threading
from collections import defaultdict
input = sys.stdin.readline
def rl(type):
	return list(map(type,input().split()))


def main():
	MOD = 10**9 + 7
	def create_graph(n):
		graph = defaultdict(list)
		for _ in range(n):
			x,y = map(int,input().split())
			graph[x].append(y)
			graph[y].append(x)
		return graph

	def walk(u,v):
		s = 1
		for i in graph[v]:
			if i != u:
				c = walk(v,i)
				w.append(c*(n-c))
				s += c
		return s

	for _ in range(int(input())):
		n = int(input())
		graph = create_graph(n-1)
		k = int(input())
		arr = rl(int)
		# print(arr)
		if k < n - 1:
			arr += [1] * (n-1-k)
			arr.sort(reverse = True)
		else:
			arr.sort(reverse=True)
			while len(arr) != (n-1):
				temp = (arr[0]*arr[1])%MOD
				arr.pop(0)
				arr[0] = temp
			# print("Esre")
			# arr.sort(reverse=True)
			# p = len(arr)
			# i = 1
			# temp = 0
			# parr  = []
			# ptemp = arr[0]
			# while p != (n-1):
			# 	temp = (arr[i] * ptemp)
			# 	ptemp = temp
			# 	i += 1
			# 	p -= 1
			# parr.append(temp)
			# for i in range(i,len(arr)):
			# 	parr.append(arr[i])
			# arr = parr

		# print(arr)
		w = []
		walk(0,1)
		w.sort(reverse=True)
		result = 0
		for i in range(n-1):
			result +=  ((w[i] % MOD) * (arr[i] % MOD)) % MOD
			result %= MOD
		print(result)

sys.setrecursionlimit(10**6)
threading.stack_size(10**8)
t = threading.Thread(target=main)
t.start()
t.join()