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 functools import reduce

class Node:
  def __init__(self, val: int):
    self.val=val
    self.count=1

def dfs(root, tree) -> int:
  n = len(tree) -1
  visited = [False]*(n+1) 
  completed= [False]*(n+1)
  stack = [root]
  while len(stack)>0:
    node = stack[-1]
    if visited[node.val]:
      node.count += sum(child.count for child in tree[node.val] if completed[child.val])
      stack.pop()
      completed[node.val] = True
    else:
      for child in tree[node.val]:
        if not visited[child.val]:
          stack.append(child)
      visited[node.val] = True

t = int(input())
while t>0:
  n = int(input())
  tree = [[] for i in range(0,n+1)]
  nodes = [Node(val) for val in range(0,n+1)]
  for i in range(0,n-1):
    u,v = map(int, input().split())
    tree[u].append(nodes[v])
    tree[v].append(nodes[u])
  m = int(input())
  factors = list(map(int, input().split()))
  m = len(factors)
  factors.sort(reverse=True)
  dfs(nodes[1],tree)
  edge_multipliers = [node.count*(n-node.count) for node in nodes if node.count!=n and node.val!=0]
  edge_multipliers.sort(reverse=True)
  result = 0
  mod = int(1e9+7)
  result = 0
  if m <=n-1:
    for i in range(0,n-1):
      p = edge_multipliers[i]%mod 
      if i<m:
        p = (factors[i]*p)%mod
      result = (result + p)%mod
  else:
    p0 = 1
    for i in range(0,m-n+2):
      p0 = (p0*factors[i])%mod
    factors = [p0] + factors[m-n+2:m]
    for i in range(0,n-1):
      result = (result + (factors[i]*(edge_multipliers[i]%mod))%mod)%mod
  print(result)
  t -= 1
