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.
from sys import stdin
input=lambda : stdin.readline().strip()
lin=lambda :list(map(int,input().split()))
iin=lambda :int(input())
main=lambda :map(int,input().split())
from math import ceil,sqrt,factorial,log
from collections import deque
from bisect import bisect_left
def gcd(a,b):
	a,b=max(a,b),min(a,b)
	while a%b!=0:
		a,b=b,a%b
	return b
mod=998244353
mod=1000000007
def dfs(tree,n):
	child={i:1 for i in range(n)}
	parent={i:-1 for i in range(n)}
	vis=[0]*n
	s=deque([0])
	q=[]
	while s:
		x=s.popleft()
		q.append(x)
		vis[x]=1
		for i in tree[x]:
			if vis[i]==0:
				s.append(i)
				parent[i]=x
	# print(q)
	for i in range(n-1,0,-1):
		child[parent[q[i]]]+=child[q[i]]
		# print(i)
		# print(q[i],i)

	# print(child)
	l=[]
	for i in range(1,n):
		l.append(child[i]*(n-child[i]))
	# print(l)
	l.sort()
	return l
def solve(we):
	n=iin()
	tree=[[] for i in range(n)]
	for i in range(n-1):
		u,v=main()
		u-=1
		v-=1
		tree[u].append(v)
		tree[v].append(u)
	# print(tree)
	z=dfs(tree,n)
	m=iin()
	l=lin()
	l.sort()
	if m<n-1:
		l=[1 for i in range(n-m-1)]+l
	else:
		t=1
		for i in range(n-2,m):
			t*=l[i]
			t=t%mod
		l=l[:n-2]+[t]
	ans=0
	for i in range(n-1):
		ans=(ans+(z[i]*l[i])%mod)%mod
	print(ans)
qwe=1
qwe=iin()
for _ in range(qwe):
	solve(_+1)