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.
from __future__ import division, print_function
import sys
if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

import os, sys, bisect, copy
from collections import defaultdict, Counter, deque
#from functools import lru_cache   #use @lru_cache(None)
if os.path.exists('in.txt'): sys.stdin=open('in.txt','r')
if os.path.exists('out.txt'): sys.stdout=open('out.txt', 'w')
#
def input(): return sys.stdin.readline()
def mapi(arg=0): return map(int if arg==0 else str,input().split())
#------------------------------------------------------------------

import sys
sys.setrecursionlimit(10**5+10)
from types import GeneratorType
#use:- put @bootstrap overrecursive function
def bootstrap(func, stack=[]):
    def wrapped_function(*args, **kwargs):
        if stack:
            return func(*args, **kwargs)
        else:
            call = func(*args, **kwargs)
            while True:
                if type(call) is GeneratorType:
                    stack.append(call)
                    call = next(call)
                else:
                    stack.pop()
                    if not stack:
                        break
                    call = stack[-1].send(call)
            return call
    return wrapped_function


mod = int(1e9+7)
for _ in range(int(input())):
    n = int(input())
    gr = defaultdict(list)
    for i in range(n-1):
        x,y = mapi()
        gr[x].append(y)
        gr[y].append(x)
    m = int(input())
    p = list(mapi())
    # p = p+[1]*(n-1-len(p))
    while len(p)<n-1:
        p.append(1)
    p.sort()
    for i in range(n-1,len(p),1):
        p[n-2] = p[n-2]*p[i]%mod
    p = p[:n-1]
    p = p[::-1]
    sz = [1]*(n+1)
    @bootstrap
    def dfs(s,par):
        for i in gr[s]:
            if i==par: continue
            yield dfs(i,s)
            sz[s] +=sz[i]
        yield 0
    dfs(1,-1)
    count = []
    for i in range(1,n+1):
        count.append(sz[i]*(n-sz[i]))
    count.sort(reverse=True)
    res = 0
    for i in range(n-1):
        res+=p[i]*count[i]%mod
        res%=mod
    print(res)
