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.
# -*- coding: utf-8 -*-

import sys
# sys.setrecursionlimit(2 * 10**5)
# buff_readline = sys.stdin.buffer.readline
buff_readline = sys.stdin.readline
readline = sys.stdin.readline

INF = 2**62-1


def read_int():
    return int(buff_readline())


def read_int_n():
    return list(map(int, buff_readline().split()))


def read_float():
    return float(buff_readline())


def read_float_n():
    return list(map(float, buff_readline().split()))


def read_str():
    return readline().strip()


def read_str_n():
    return readline().strip().split()

def error_print(*args):
    print(*args, file=sys.stderr)


def mt(f):
    import time

    def wrap(*args, **kwargs):
        s = time.time()
        ret = f(*args, **kwargs)
        e = time.time()

        error_print(e - s, 'sec')
        return ret

    return wrap

from collections import defaultdict, Counter

# @mt
def slv(N, UV, M, P):
    g = defaultdict(list)
    for u, v in UV:
        g[u].append(v)
        g[v].append(u)

    np = Counter()

    # def dfs(u, r):
    #     n = 0
    #     vm = {}
    #     for v in g[u]:
    #         if v == r:
    #             continue
    #         m = dfs(v, u)
    #         n += m
    #         vm[v] = m

    #     for v, m in vm.items():
    #         np[(min(u, v), max(u, v))] = (N-m) * m

    #     return n + 1
    # dfs(1, -1)

    s = [(u, -1)]
    o = []
    while s:
        u, r = s.pop()
        o.append((u, r))
        for v in g[u]:
            if v == r:
                continue
            s.append((v, u))

    o.reverse()
    cn = [0] * (N+1)
    for u, r in o:
        n = 0
        vm = {}
        for v in g[u]:
            if v == r:
                continue
            n += cn[v]
            vm[v] = cn[v]
        cn[u] = n + 1
        for v, m in vm.items():
            np[(min(u, v), max(u, v))] = (N-m) * m

    np = list(np.values())

    np.sort()
    P.sort()
    M = 10**9 + 7
    while len(P) > len(np):
        p = P.pop()
        P[-1] *= p
        P[-1] %= M
    if len(P) < len(np):
        P = [1] * (len(np) - len(P)) + P

    ans = 0

    for v, p in zip(np, P):
        ans += v * p
        ans %= M

    return ans



def main():
    for _ in range(read_int()):
        N = read_int()
        UV = [read_int_n() for _ in range(N-1)]
        M = read_int()
        P = read_int_n()
        print(slv(N, UV, M, P))


    # N = 10**5
    # UV = [[i, i+1] for i in range(1, N)]
    # M = 2
    # P = [2, 3]
    # print(slv(N, UV, M, P))

if __name__ == '__main__':
    main()
