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.
def tInput():
    from sys import stdin
    input=stdin.readline
tInput()
 
 
def A():
    for _ in range(int(input())):
        n,k=map(int,input().split())
        if n>=k: print(int(n%2!=k%2))
        else: print(k-n)
def B():
    for _ in range(int(input())):
        x1,y1,z1=map(int,input().split())
        x2,y2,z2=map(int,input().split())
        tot=0
        p=min(x1,z2)
        x1 -= p
        z2 -= p        
        tot+=2*min(z1,y2)
        p=min(z1,y2)
        z1-=p
        y2-=p
        if z2 > 0:
            z2 -= min(z2,z1)
            tot-= 2*min(y1,z2)
        print(tot)
def C():
    t=int(input())
    for _ in range(t):
        n=int(input())
        a=list(map(int,input().split()))
        sa=sorted(a)
        m=min(a)
        w=True
        for i in range(n):
            if a[i]%m != 0:
                if a[i] != sa[i]:
                    w=False
                    break
        if w: print("YES")
        else: print("NO")
def D():
    t=int(input())
    from collections import deque
    for _ in range(t):
        n=int(input())
        graph=[set() for i in range(n+1)]
        graph[1].add(0)
        queue=deque()
        #count the number of nodes below and above each edge. multiply these together and sort
        numChildren=[0]*(n+1)
        
        for i in range(n-1):
            u,v=map(int,input().split())
            graph[u].add(v)
            graph[v].add(u)
        for i in range(1,n+1):
            if len(graph[i]) == 1 and i != 1:
                queue.append(i)
        while len(queue) > 0:
            curr=queue.popleft()
            if curr == 1:
                break
            parent=graph[curr].pop()
            numChildren[curr] += 1
            numChildren[parent] += numChildren[curr]
            graph[parent].remove(curr)
            if len(graph[parent]) == 1:
                queue.append(parent)
 
        mults=[0]*(n-1)
        for i in range(2,n+1):
            mults[i-2]=numChildren[i]*(n - numChildren[i])
        mults.sort()
        mod=10**9+7
        m=int(input())
        p=list(map(int,input().split()))
        if m <= n - 1:
            p = p + [1]*(n-m-1)
            p.sort()
        else:
            p.sort()
            for i in range(n - 1, m):
                p[n-2] = (p[n-2] * p[i]) % mod
        tot=0
        for i in range(n-1):
            tot = (tot + p[i]*mults[i]) % mod
        print(tot)
D()
     