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.
#write dfs recursively

import sys

def input():
	return sys.stdin.readline().rstrip()

def input_split():
	return [int(i) for i in input().split()]

testCases = int(input())
answers = []
sys.setrecursionlimit(10**5) 
#dfs returns size
# def dfs(current, edges, freqs, touched):
# 	global n
# 	ans = 1
# 	touched[current] = True
# 	for child in edges[current]:
# 		if not touched[child]:
# 			subtree = dfs(child, edges, freqs, touched) 
# 			freqs[(min(current, child), max(current, child))] = subtree * (n - subtree)
# 			ans += subtree
# 	return ans


for _ in range(testCases):
	#take input
	n = int(input())
	edges = [[] for i in range(n)]
	freqs = {}
	for _ in range(n-1):
		u, v = input_split()
		u -= 1
		v -= 1
		edges[u].append(v)
		edges[v].append(u)
		freqs[((min(u,v)), max(u,v))] = 0

	m = int(input())
	primes = input_split()
	MOD = 10**9 + 7

	primes.sort()

	#now i want to do a dfs
	# for i in range()
	#time for dfs

	touched = [False for i in range(n)]
	pending = [0]
	size = 1
	all_parents = [-1 for i in range(n)]

	all_sizes = [-1 for i in range(n)]
	descendants = [[] for i in range(n)]
	while(len(pending) != 0):
		current = pending.pop()
		if current == -1:
			#store size info
			parent = pending.pop()
			size = 1
			for desc in descendants[parent]:
				size += all_sizes[desc]

			all_sizes[parent] = size

			u = all_parents[parent]
			v = parent
			if u == -1:
				continue
				#means ki khatam hogaya
			mini = min(u,v)
			maxi = max(u,v)
			freqs[(mini, maxi)] = size * (n - size)
			# size = 1
			continue

		touched[current] = True
		pending.append(current)
		pending.append(-1)
		for child in edges[current]:
			if not touched[child]:
				descendants[current].append(child)
				all_parents[child] = current
				pending.append(child)


	# dfs(0, edges, freqs, touched)

	e_f = list(freqs.values())
	e_f.sort()

	if len(primes) <= len(e_f):
		primes = [1]*(len(e_f) - len(primes)) + primes
	else:
		front = primes[:len(e_f)]
		p = 1
		for i in range(len(e_f) , len(primes)):
			p = (p*primes[i])%MOD

		front[-1] *= p 
		primes = front

	ans = 0
	for i in range(len(primes)):
		ans = (ans + primes[i]*e_f[i])%MOD

	answers.append(ans)

print(*answers, sep = '\n')
