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,math,threading,heapq
sys.setrecursionlimit(10**7)
threading.stack_size(10**8)

def solve():
    def get_int(): # returns a list of integers after taking an input
        return list(map(int,sys.stdin.readline().strip().split()))
    

    def dfs(adj_list,imp_edges,v_list,n,u=1):
        child=1
        for v in adj_list[u]:
            if v_list[v]==False:
                v_list[v]=True
                x=dfs(adj_list,imp_edges,v_list,n,v)
                heapq.heappush(imp_edges,(-(x*(n-x)),u,v))
                child+=x
        # print(u,child)
        return child

    if __name__ == "__main__":
        t=int(input())
        mod=10**9+7
        for _ in range(t):
            N=int(input().strip())
            adj_list=[[] for _ in range(N+1)] #Stores the adjacency list
            adj_list[0]=None
            for _ in range(N-1):
                a,b=get_int()
                adj_list[a].append(b) 
                adj_list[b].append(a)
            imp_edges=[]
            heapq.heapify(imp_edges)
            visited=[False]*(N+1)
            visited[1]=True
            dfs(adj_list,imp_edges,visited,N)
            # print(imp_edges)
            int(input())
            fctrs=get_int()
            fctrs.sort()
            # print(fctrs)

            ans=0
            while len(fctrs)>len(imp_edges):
                lst_fctr=(fctrs.pop()*fctrs.pop())%mod
                fctrs.append(lst_fctr)
            while len(imp_edges)>0:
                ans=((ans%mod)+(abs(imp_edges[0][0])*(1 if len(fctrs)==0 else fctrs.pop()))%mod)%mod
                heapq.heappop(imp_edges)
            # print()
            print(ans)

threading.Thread(target=solve).start()

