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
from collections import deque

input=sys.stdin.readline

for _ in range(int(input())):
    n=int(input())
    edge=[[] for i in range(n)]
    for i in range(n-1):
        u,v=map(int,input().split())
        edge[u-1].append(v-1)
        edge[v-1].append(u-1)
    m=int(input())
    p=list(map(int,input().split()))

    size=[1 for i in range(n)]
    parent=[-1 for i in range(n)]
    res=[]
    deq=deque([(0,-1)])
    while deq:
        v,pv=deq.popleft()
        res.append(v)
        for nv in edge[v]:
            if nv!=pv:
                parent[nv]=v
                deq.append((nv,v))

    res=res[::-1]
    for v in res:
        for nv in edge[v]:
            if nv!=parent[v]:
                size[v]+=size[nv]

    coef=[]
    for v in res:
        for nv in edge[v]:
            if nv!=parent[v]:
                c=size[nv]*(n-size[nv])
                coef.append(c)
    mod=10**9+7
    #print(coef)
    #print(p)
    #print(size)
    if m<n-1:
        res=0
        coef.sort(reverse=True)
        p.sort(reverse=True)
        for i in range(m):
            res+=coef[i]*p[i]
            res%=mod
        for i in range(m,n-1):
            res+=coef[i]
            res%=mod
        print(res)
    else:
        res=0
        coef.sort()
        p.sort()
        for i in range(n-2):
            res+=coef[i]*p[i]
            res%=mod
        tmp=coef[-1]
        for i in range(n-2,m):
            tmp*=p[i]
            tmp%=mod
        res=(res+tmp)%mod
        print(res)
