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 types import GeneratorType


# Pajenegod's Infinite Recursion trick
def recursive(f, stack=[]):
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to

    return wrappedfunc


from sys import stdin


def inp():
    return stdin.buffer.readline().rstrip().decode('ascii')


def itg():
    return int(stdin.buffer.readline())


def mpint():
    return map(int, stdin.buffer.readline().split())


# ############################## import
# from ACgenerator.Calculator import tree_edge_descendant
def tree_edge_descendant(graph, edge_time_count=False, root=0):
    """
    :return: a list for the number of vertices under each edge

    if edge_time_count:
    :return a list for the sum of times that walking on each edge
    the walk mean the path of any pair (u, v) and (u < v)
    """

    edges = len(graph) - 1
    descendant_count = []
    stack_par = [[root]]
    stack_val = [[0]]
    while len(descendant_count) != edges:
        if not stack_par[-1]:
            del stack_par[-1], stack_val[-1]
            value = stack_val[-1].pop() + 1
            descendant_count.append(value)
            stack_val[-2][-1] += value
            if not stack_val:
                del stack_val[-1]
            continue
        vertex = stack_par[-1].pop()
        stack_par.append(list(graph[vertex]))
        stack_val.append([0] * len(graph[vertex]))

    if edge_time_count:
        return list(map(lambda x: x * (len(graph) - x), descendant_count))
    return descendant_count


# from ACgenerator.Tools import to_tree
def to_tree(graph, root=0):
    """
    graph: undirected graph (adjacency list)
    :return directed graph that parent -> children
    """
    graph[:] = map(set, graph)

    @recursive
    def dfs(v):
        for e in graph[v]:
            graph[e].remove(v)
            yield dfs(e)
        yield

    dfs(root)
    graph[:] = map(list, graph)


# ############################## main

from functools import reduce

MOD = int(1e9) + 7


def solve():
    g = []
    for __ in range(itg()):
        n = itg()
        g = [list() for _ in range(n)]
        for _ in range(n - 1):
            u, v = mpint()
            g[u - 1].append(v - 1)
            g[v - 1].append(u - 1)
        to_tree(g)
        m = itg()
        factors = sorted(mpint())
        count = sorted(tree_edge_descendant(g, True))
        n -= 1
        if n >= m:
            factors = ([1] * (n - m)) + factors
        else:  # n < m
            while len(factors) != n:
                factors[-1] = factors[-2] * factors.pop() % MOD

        print(
            reduce(
                lambda a, b: (a + b) % MOD,
                (time * value % MOD for time, value in zip(count, factors)),
                0
            )
        )


if __name__ == '__main__':
    solve()
# Please check!
