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.
######################################################
############Created by Devesh Kumar###################
#############devesh1102@gmail.com####################
##########For CodeForces(Devesh1102)#################
#####################2020#############################
######################################################
import sys
input = sys.stdin.readline
 
# import sys
import heapq 
import copy
import math
import decimal
 
# import sys.stdout.flush as flush
# from decimal import *
#heapq.heapify(li) 
#
#heapq.heappush(li,4) 
#
#heapq.heappop(li)
#
# &    Bitwise AND Operator    10 & 7 = 2
# |    Bitwise OR Operator    10 | 7 = 15
# ^    Bitwise XOR Operator    10 ^ 7 = 13
 
# <<    Bitwise Left Shift operator    10<<2 = 40
# >>    Bitwise Right Shift Operator
# '''############ ---- Input Functions ---- #######Start#####'''
 
 
def inp():
    return(int(input()))
def inlt(): 
    return(list(map(int,input().split())))
def insr():
    s = input()
    return(list(s[:len(s) - 1]))
def insr2():
    s = input()
    return((s[:len(s) - 1]))
def invr():
    return(map(int,input().split()))
############ ---- Input Functions ---- #######End
# #####   
 

ans = 0
def pr_list(a):
    print( *a , sep=" ")
def main():
    tests =  inp()
    # tests = 1
    mod = 1000000007
    limit = 10**18
    ans = 0
    stack = []
    hashm = {}
    arr = []
    heapq.heapify(arr) 

    for test in range(tests):
        n = inp()
        grid = [{} for i in range(n+1)]
        for i in range(n-1):
            [a,b] = inlt()
            grid[a][b] =1
            grid[b][a] = 1
        m = inp()
        p = inlt()
        
        p.sort(reverse = True)
        if m > n -1:
            # print(p,len(p))
            d = 1
            for i in range(m-n + 2):
                a = p.pop(0)
                d = (d*a)%mod
            p = [d] + p
            # print(p,len(p))
        m = len(p)
        parent = [0 for i in range(n+1)]
        nodes = [0 for i in range(n+1)]
        times = []
        queue = [1]
        done = [0 for i in range(n+1)]
        done[1] =1
        # print(grid)
        while(queue!= []):
            # print(queue,"here",times)
            curr = queue[-1]
            flag = 0
            for i in grid[curr]:
                if i != parent[curr]:
                    if curr in grid[i]:
                        del grid[i][curr]
                    parent[i] = curr 
                if done[i] == 0:
                    done[i]= 1
                    flag = 1
                    queue.append(i)
                
            if flag == 1:
                continue 
            curr = queue.pop()
            reach = 1
            for i in grid[curr]:
                up = (n - nodes[i])* nodes[i]
                # down =
                times.append(up)
                reach = reach + nodes[i]
            nodes[curr] = reach
            

        times.sort(reverse = True)
        # print(times)
        # print(grid)

        ans = 0
        for i in range(len(times)):
            # print(p[i])
            if i < m:
                ans =( ans + times[i]*p[i]  )%mod
            else:
                ans =(ans + times[i]*1)%mod
        print(ans)


if __name__== "__main__":
    main()
