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 sys
input=sys.stdin.buffer.readline
inin=lambda: int(input())
inar=lambda: list(map(int,input().split()))
inst=lambda: input().decode().rstrip('\n\r')
#INF=float('inf'); from math import gcd 
#from collections import deque as que, defaultdict as vector, Counter
#from bisect import bisect
#from heapq import heapify, heappush as hpush, heappop as hpop
from types import GeneratorType
def recursive(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
gr=[]
childs=[]
n=0
@recursive
def dfs(node,par):
    global gr,cost,n,childs
    su=1
    for i in gr[node]:
        if i!=par:
            yield dfs(i,node)
            su+=childs[i]
            cost.append(childs[i]*(n-childs[i]))
    childs[node]=su
    yield
mod=1000000007
_T_=inin()
for _t_ in range(_T_):
    #print(_t_)
    n=inin()
    gr=[[] for i in range(n+1)]
    childs=[1 for i in range(n+1)]
    cost=[]
    for _ in range(n-1):
        u,v=inar()
        gr[u].append(v)
        gr[v].append(u)
    m=inin()
    cm=inar()
    cm.sort()
    dfs(1,0)
    #print(childs)
    cost.sort()
    #print(cost)
    ll=len(cost)
    if ll>=m:
        cm=[1 for i in range(ll-m)]+cm 
    else:
        while(len(cm)!=ll):
            cm[-2]*=cm[-1]
            cm[-2]%=mod #dont underestimate this line
            cm.pop()
    s=0
    #print(cm,cost)
    for i in range(ll):
        s+=((cm[i]%mod)*(cost[i]%mod))%mod
        s%=mod
    print(s)
