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.
import os
import sys
from io import BytesIO, IOBase

_str = str
BUFSIZE = 8192


def str(x=b''):
    return x if type(x) is bytes else _str(x).encode()


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = 'x' in file.mode or 'r' not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b'\n') + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode('ascii'))
        self.read = lambda: self.buffer.read().decode('ascii')
        self.readline = lambda: self.buffer.readline().decode('ascii')


def inp():
    return sys.stdin.readline().rstrip()


def mpint():
    return map(int, inp().split(' '))


def itg():
    return int(inp())


# ############################## import
# from ACgenerator.Y_Testing import get_code
def tree_bfs(tree, flat=True, start=0) -> list:
    stack = [start]
    result = []
    while stack:
        new_stack = []
        if flat:
            result.extend(stack)
        else:
            result.append(stack)
        for node in stack:
            new_stack.extend(tree[node])
        stack = new_stack
    return result


def tree_postorder(tree, flat=True, root=0):
    """ children -> parent """
    return tree_bfs(tree, flat, root)[::-1]


def get_parent_list(tree):
    result = [0] * len(tree)
    for parent, children in enumerate(tree):
        for child in children:
            result[child] = parent
    return result


def to_tree(graph, root=0):
    """
    graph: [{child, ...}, ...]  (undirected)
    :return directed graph that parent -> children
    """
    stack = [[root]]
    while stack:
        if not stack[-1]:
            del stack[-1]
            continue
        parent = stack[-1].pop()
        for child in graph[parent]:
            graph[child].discard(parent)
        stack.append(list(graph[parent]))


def edge_descendant(tree, edge_time_count=False):
    """
    :return: [{child: num, ...}, ...]
    the number of vertices under each edge

    if edge_time_count:
    the num will translate to the sum of times that walking on the edge
    the walk mean the path of any pair (u, v) and (u < v)
    """
    vertices = len(tree)
    results = [{} for _ in range(vertices)]
    parent_list = get_parent_list(tree)
    for child in tree_postorder(tree):
        parent = parent_list[child]
        if parent != child:
            results[parent][child] = sum(results[child].values()) + 1

    if edge_time_count:
        for parent in results:
            for child in parent:
                num = parent[child]
                parent[child] = num * (vertices - num)
    return results


# ############################## main
from itertools import chain
from functools import reduce

MOD = 10 ** 9 + 7


def solve():
    n = itg()
    tree = [set() for _ in range(n)]
    for _ in range(n - 1):
        u, v = mpint()
        tree[u - 1].add(v - 1)
        tree[v - 1].add(u - 1)
    to_tree(tree)
    m = itg()
    factors = sorted(mpint())
    count = sorted(chain(*[d.values() for d in edge_descendant(tree, 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

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


def main():
    # solve()
    # print(solve())
    for _ in range(itg()):
        print(solve())
        # print("YES" if solve() else "NO")
        # print("yes" if solve() else "no")


DEBUG = 0
URL = 'https://codeforces.com/contest/1401/problem/D'

if __name__ == '__main__':
    if DEBUG == 1:
        import requests  # ImportError: cannot import name 'md5' from 'sys' (unknown location)
        from ACgenerator.Y_Test_Case_Runner import TestCaseRunner

        runner = TestCaseRunner(main, URL)
        inp = runner.input_stream
        print = runner.output_stream
        runner.checking()
    elif DEBUG == 2:
        main()
    else:
        sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
        main()
# Please check!
