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.
# NOT MY CODE
# https://codeforces.com/contest/1324/submission/73179914
 
## PYRIVAL BOOTSTRAP
# https://github.com/cheran-senthil/PyRival/blob/master/pyrival/misc/bootstrap.py
# This decorator allows for recursion without actually doing recursion
from types import GeneratorType
 
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


import sys,math
input=sys.stdin.readline
t=int(input())
mod = (10**9) + 7

count1 = None

@bootstrap
def numberOfNodes(s, e):  
  
    count1[s] = 1
    for u in graph[s]:  
          
        # Condition to omit reverse path  
        # path from children to parent  
        if u == e:  
            continue
          
        # recursive call for DFS  
        yield numberOfNodes(u, s)    
        count1[s] += count1[u] 
    yield

for r in range(t):
    n = int(input())
    graph = []
    edges = []
    for i in range(n):
        graph.append([])
    for i in range(n-1):
        u, v = map(int,input().split())
        u -= 1
        v -=1
        graph[u].append(v)
        graph[v].append(u)
        edges.append([u,v])
    m = int(input())
    p = list(map(int,input().split()))
    count1 = [0] * n
    numberOfNodes(0,-1)
    # print(count1)
    edgesFrequency = []
    for i in range(1,n):
        edgesFrequency.append(count1[i] * (n - count1[i]))
    # print(edgesFrequency)
    p.sort(reverse = True)
    edgesFrequency.sort(reverse = True)
    if (n-1) > m:
        diff = (n-1) - m
        for i in range(diff):
            p.append(1)
    elif m > (n-1):
        diff = m - (n-1)
        temp = 1
        for i in range(diff+1):
            temp *= p[i]
            temp %= mod
        temp2 = [temp]
        for i in range(diff+1, m):
            temp2.append(p[i])
        p = temp2
    ans = 0
    for i in range(n-1):
        ans += ((edgesFrequency[i] * p[i]) % (mod))
        ans %= (mod)
    print(ans)