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 math
from sys import stdin
g   = lambda : stdin.readline().strip()
gl  = lambda : g().split()
gil = lambda : [int(var) for var in gl()]
gfl = lambda : [float(var) for var in gl()]
gcl = lambda : list(g())
gbs = lambda : [int(var) for var in g()]
mod = int(1e9)+7
inf = float("inf")


def mul(a, b):
	return ((a%mod)*(b%mod))%mod


t, = gil()
for _ in range(t):
	n, = gil()
	adj = [[] for _ in range(n+1)]
	for _ in range(n-1):
		x, y = gil()
		adj[x].append(y)
		adj[y].append(x)


	m, = gil()
	fac = gil()
	fac.sort()
	while len(fac) >= n :
		v = fac.pop()
		fac[-1] = mul(v, fac[-1])
		# fac[-1] = v*fac[-1]
	# print(len(fac))


	dfs, vis, cnt, fin, edges = [1], [0]*(n+1), [1]*(n+1), [0]*(n+1), []

	while dfs:
		x = dfs[-1]
		if vis[x]:
			dfs.pop()
			for y in adj[x]:
				if fin[y]:
					cnt[x] += cnt[y]
					edges.append(cnt[y]*(n-cnt[y]))
			fin[x] = 1
		else:
			vis[x] = 1
			for y in adj[x]:
				if vis[y] == 0:
					dfs.append(y)

	edges.sort()
	# print(cnt[1:])
	# print(edges)
	# print(fac)
	ans = 0

	while edges and fac:
		ans = (mul(fac.pop(), edges.pop())+ ans)%mod

	while edges:
		ans = (ans+edges.pop())%mod

	print(ans%mod)

