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.
DISTRIBUTION_INDEX_MOD = 10**9 + 7


def compute_max_distribution_index(graph, factors_seq):
    edges_loading = _compute_edges_loading(graph)
    loadings = sorted(edges_loading.values(), reverse=True)
    edges_n = len(loadings)
    
    weights = [1] * edges_n
    factors_seq = sorted(factors_seq)
    first_factors, factors_remain = factors_seq[:edges_n], factors_seq[edges_n:]
    i_edge = 0
    while first_factors:
        weights[i_edge] *= first_factors.pop()
        weights[i_edge] %= DISTRIBUTION_INDEX_MOD
        i_edge += 1
    for factor in factors_remain:
        weights[0] *= factor
        weights[0] %= DISTRIBUTION_INDEX_MOD
    
    result = 0
    for loading, weight in zip(loadings, weights):
        loading %= DISTRIBUTION_INDEX_MOD
        result += (loading * weight) % DISTRIBUTION_INDEX_MOD
        result %= DISTRIBUTION_INDEX_MOD
    
    return result


def _compute_edges_loading(graph):
    vertices_n = len(graph)
    edges_loading = dict()
    
    leaves = [vertex for vertex in range(len(graph)) if len(graph[vertex]) == 1]
    vertices_accumulated = [None] * len(graph)
    for leave in leaves:
        vertices_accumulated[leave] = 1
        neighbors = graph[leave]
        assert len(neighbors) == 1
        next_vertex = tuple(neighbors)[0]
        edges_loading[frozenset((leave, next_vertex))] = \
            _compute_edge_loading(vertices_accumulated[leave], vertices_n)
    
    active = set().union(*(graph[leave] for leave in leaves))
    passed = set(leaves)
    active -= passed
    while len(passed) < len(graph):
        old_active = active
        active = set()
        for vertex in old_active:
            neighbors = graph[vertex]
            neighbors_not_passed = neighbors - passed
            if len(neighbors_not_passed) > 1:
                active.add(vertex)
                continue
            vertices_accumulated[vertex] = sum(
                vertices_accumulated[neighbor] for neighbor in neighbors if neighbor in passed
            ) + 1
            for neighbor in neighbors_not_passed:
                edges_loading[frozenset((vertex, neighbor))] = \
                    _compute_edge_loading(vertices_accumulated[vertex], vertices_n)
            passed.add(vertex)
            active.update(neighbors_not_passed)
        active = active - passed
    
    return edges_loading


def _compute_edge_loading(accumulated_n, vertices_n):
    return accumulated_n * (vertices_n - accumulated_n)


def main():
    t = int(input())
    for i in range(t):
        n = int(input())
        graph = tuple(set() for i in range(n))
        for i_edge in range(n-1):
            v1, v2 = map(int, input().split())
            v1, v2 = v1-1, v2-1
            graph[v1].add(v2)
            graph[v2].add(v1)
        m = int(input())
        factors_seq = tuple(map(int, input().split()))
        result = compute_max_distribution_index(graph, factors_seq)
        print(result)


if __name__ == '__main__':
    main()
