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 math,sys
#from itertools import permutations, combinations;import heapq,random;
from collections import defaultdict,deque
import bisect as bi
def yes():print('YES')
def no():print('NO')
def I():return (int(sys.stdin.readline()))
def In():return(map(int,sys.stdin.readline().split()))
def Sn():return sys.stdin.readline().strip()
#sys.setrecursionlimit(1500)
def dict(a):
    d={} 
    for x in a:
        if d.get(x,-1)!=-1:
            d[x]+=1
        else:
            d[x]=1
    return d
def find_gt(a, x):
    'Find leftmost value greater than x'
    i = bi.bisect_left(a, x)
    if i != len(a):
        return i
    else:            
        return -1
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
@bootstrap
def dfs(node,d):
    if visit[node]:
        return
    visit[node]=True
    su=0
    for x in d[node]:
        if not visit[x]:
            su+=yield(dfs(x,d))
    dp[node]=su+1
    yield (su+1)
    
def main():
    try:
        global visit,dp
        n=I()
        d=defaultdict(list)
        for x in range(n-1):
            a,b=In()
            d[a].append(b)
            d[b].append(a)
        m=I()
        primes=list(In())
        primes.sort()
        dp=[0]*(n+1)
        visit=[False] *(n+1)
        dfs(1,d)
        time=[]
        for i in range(1,n+1):
            z=(n-dp[i])*dp[i]
            if z:
                time.append(z)
        time.sort()
        ans=0
        if m>len(time):
            j=0
            for i in range(len(time)-1):
                ans=ans%P+(time[i]%P*primes[j])%P
                ans%=P
                j+=1
            an=1
            for i in range(j,m):
                an*=primes[i]
                an%=P
            ans=ans%P+(time[-1]%P*an)%P
            print(ans%P)
        else:
            te=[]
            for i in range(len(time)-m):
                te.append(1)
            te+=primes
            for i in range(len(time)):
                ans+=(time[i]%P*te[i])%P
                ans%=P
            print(ans%P)
    except:
        pass
        
M = 998244353
P = 1000000007
 
if __name__ == '__main__':
    for _ in range(I()):main()
    # for _ in range(1):main()
#        *******************    All The Best    *******************   #