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 io, os
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline

from collections import defaultdict, deque

t = int(input())
MOD = 10**9 + 7

for _ in range(t):
    n = int(input())
    adj = defaultdict(list)
    edges_list = []
    
    for i in range(n-1):
        u, v = map(int, input().split())
        adj[u].append(v)
        adj[v].append(u)
        edges_list.append((u, v))
        
    m = int(input())
    a = list(map(int, input().split()))
    a.sort(reverse=True)
    
    size = [0]*(n+1)
    parents = [0]*(n+1)

    states = []
    q = deque([[1, -1]])
    while q:
        e, p = q.popleft()
        states.append(e)
        parents[e] = p

        for i in adj[e]:
            if i != p:
                q.append([i, e])

    for i in range(n, 0, -1):
        for j in adj[states[i-1]]:
            if states[i-1] == parents[j]:
                size[states[i-1]] += 1+size[j]
    
    weights = None
    if m >= n:
        first = a[0]
        second = deque(a[1:])
        
        while len(second) != n-2:
            first = (first*second.popleft())%MOD
            
        weights = [first] + list(second)
        
    else:
        weights = a.copy()
        while len(weights) != n-1:
            weights.append(1)
        
    edge_bckt = []
    
    for i in edges_list:
        query = min(size[i[0]], size[i[1]])
        edge_bckt.append((query+1)*(n-query-1))
        
    edge_bckt.sort(reverse=True)
    prod = 0
    for i in range(len(edge_bckt)):
        prod = (prod+edge_bckt[i]*weights[i])%MOD
        
    print(prod)