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 collections import defaultdict
import sys
from types import GeneratorType

MOD = (10 ** 9) + 7


def bootstrap(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


def rl():
    return [int(w) for w in sys.stdin.readline().split()]


@bootstrap
def number_of_nodes(s, e):
    counts[s] = 1
    for u in edges[s]:
        if u == e:
            continue
        yield number_of_nodes(u, s)
        counts[s] += counts[u]
    yield


for _ in range(rl()[0]):
    (n,) = rl()
    edges = defaultdict(set)

    for _ in range(n - 1):
        u, v = rl()
        edges[u - 1].add(v - 1)
        edges[v - 1].add(u - 1)

    (m,) = rl()
    p = rl()
    p.sort(reverse=True)

    counts = [0] * n
    number_of_nodes(0, -1)

    edge_frequency = []
    for i in range(1, n):
        edge_frequency.append(counts[i] * (n - counts[i]))
    edge_frequency.sort(reverse=True)

    while m < n - 1:
        p.append(1)
        m += 1

    if m > n - 1:
        d = m - (n - 1) + 1
        a = 1
        for i in range(d):
            a *= p[i]
            a %= MOD
        p = [a] + p[d:]

    #   print(n - 1, len(p))
    f = 0
    for i in range(n - 1):
        f = (f + edge_frequency[i] * p[i]) % MOD

    print(f)
