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
from collections import deque
import bisect
#input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
RL = lambda : list(map(int, input().split(' ')))
#Level or height
def Levels(graph, V):      #graph or tree
    level = [None]*V
    dfs = []    #required for dfs
    st = [0]    #stack
    while st:
        node = st.pop()
        dfs.append(node)
        for child in graph[node]:
            if child==par[node]:
                continue
            par[child]=node
            st.append(child)
    return dfs
##########################
m = 10**9+7
def addEdge(a, b):  
    global adj
    adj[a].append(b)  
    adj[b].append(a)  
  
T = int(input())
for _ in range(T):
    N = int(input())
    count1 = [1] * (N)  
    par = [-1]*N
    adj = [[] for i in range(N)]
    
    for i in range(N-1):
        x, y = map(int,input().split())
        addEdge(x-1,y-1)
    pp = int(input())
    primes = RL()
    dfs = Levels(adj,N)
    #Counting descendants
    for b in dfs[::-1]:
        if b == 0:
            break
        count1[par[b]]+=count1[b]
    ################################
    ma = []
    for i in range(N):
        ma.append((N-count1[i])*count1[i])
    ma.sort(reverse=True)
    primes.sort(reverse = True)
    ans = 0
    for i in range(1,max(0,len(primes)-(N-1))+1):
        primes[0] = ((primes[0])%m*(primes[i]%m))%m
        primes[i] = 0
    primes.sort(reverse=True)
    for i in range(len(ma)):
        if i>=len(primes):
            ans=((ans%m)+(ma[i]%m))%m
        else:
            ans=((ans%m)+((ma[i]%m)*(primes[i]%m))%m)%m
    # mul = 1
    
        


    # for i in range(len(primes)):
    #     mul  = ((mul%m)*(primes[i]%m))%m
    # ans = ((ma[0]%m)*(mul%m))%m
    # print(mul,ans,ma)
    # for i in range(1,len(ma)):
    #     ans=((ans%m)+(ma[i]%m))%m
    print(ans%m)



