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 sys
import threading
from sys import stdin, stdout

sys.setrecursionlimit(10**9)
threading.stack_size(16*2048*2048)


def maximum_distributed_tree(n, uv_dic, p_a, edg):
    MOD = 10**9 + 7
    dfs(n, uv_dic, 1, -1, edg)
    edg.sort(reverse=True)
    p_a.sort(reverse=True)

    res = 0
    if len(p_a) - n + 1 > 0:
        for i in range(1, len(p_a) - n + 2):
            p_a[i] *= p_a[i-1]
            p_a[i] %= MOD
        del p_a[:(len(p_a) - n + 1)]

    for i in range(len(p_a)):
        res += edg[i]*p_a[i]
        res %= MOD

    for i in range(len(p_a), n-1):
        res += edg[i]
        res %= MOD

    return res


def dfs(n, uv_dic, node, parent, edg):

    cnt = 1
    nxt_a = uv_dic[node]
    for nxt in nxt_a:
        if nxt[1] != parent:
            nxtCnt = dfs(n, uv_dic, nxt[1], node, edg)
            edg[nxt[0]] = nxtCnt * (n - nxtCnt)
            cnt += nxtCnt

    return cnt


def solve():
    try:
        t = int(stdin.readline())
        for _ in range(t):
            n = int(stdin.readline())
            uv_dic = {}
            edg = [0 for _ in range(n-1)]
            for i in range(n-1):
                u, v = map(int, stdin.readline().split())
                if u not in uv_dic:
                    uv_dic[u] = []
                if v not in uv_dic:
                    uv_dic[v] = []
                uv_dic[u].append([i, v])
                uv_dic[v].append([i, u])

            m = int(stdin.readline())
            p_a = list(map(int, stdin.readline().split()))

            ans = maximum_distributed_tree(n, uv_dic, p_a, edg)
            stdout.write(str(ans) + '\n')
    except Exception as e:
        print("".join(str(e).split()))

threading.Thread(target=solve).start()
