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 threading, sys

threading.stack_size(2**26)


sys.setrecursionlimit(10**6)


def main():
	t = int(input())

	for ii in range(t):
		n = int(input())

		tree = []

		for i in range(n):
			tree.append([])

		for i in range(n-1):
			u, v = map(int, input().split())
			u-=1
			v-=1
			tree[u].append(v)
			tree[v].append(u)

		m = int(input())

		primes = list(map(int, input().split()))
		primes.sort(reverse = True)

		freq = []

		visited = [-1]*n
		
		def dfs(node):

			visited[node] = 1

			ans = 0
			for i in tree[node]:
				if visited[i] == -1:
					ans += dfs(i)

			if node == 0:
				return 0

			ans+=1
			freq.append((n-ans)*ans)
			return ans

		dfs(0)
		
		final = 0

		mod = 10**9 + 7

		freq.sort(reverse = True)
		# print(freq)
		# print(primes)
		i = 0

		if m<n:
			while i<n-1:
				if i < m:
					final+=((freq[i])%mod*(primes[i])%mod)%mod
					final = (final)%mod
					# final += freq[i]*primes[i]

				else:
					final += (freq[i])%mod
					final = (final)%mod
					# final += freq[i]

				# print(final)

				i+=1
		else:
			first = 1

			for i in range(m-n+2):
				first = (first*(primes[i])%mod)%mod
				# first = (first*(primes[i]))
				# print(i)

			# print(first)

			final = (first*(freq[0])%mod)%mod

			start = m-n+1

			for i in range(1, len(freq)):
				final += ((freq[i])%mod*(primes[start+i])%mod)%mod
				final = (final)%mod

		print(final)

threading.Thread(target=main).start()