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.
"""
NTC here
"""



def iin(): return int(input())


def lin(): return list(map(int, input().split()))

# interative DFS
def dfs(adj, start=1):
    n = len(adj)
    visited = [False]*n
    euler = []
    srt = [start]
    parent = [-1]*n
    # always start from index 1
    dp = [1]*n
    dp[0] = 0
    pv = 0
    while srt:
        v = srt.pop()
        if visited[v]:
            # dp add
            # print(v, pv, dp)
            pv = euler[-1]
            dp[v] += dp[pv]
            euler.append(v)

            continue

        euler.append(v)
        visited[v] = True
        if parent[v] != -1:
            srt.append(parent[v])
        for u in adj[v]:
            if not visited[u]:
                parent[u] = v
                srt.append(u)
        
    return dp


def main():
    md = 10**9+7
    T = iin()
    while T:
        T -= 1
        n = iin()
        adj = [[] for i in range(n+1)]
        edge = {}
        for i in range(n-1):
            j, k = lin()
            if j>k:j,k = k,j
            adj[j].append(k)
            adj[k].append(j)
            edge[(j, k)] = 1
        m = iin()
        p = lin()
        p.sort()
        dp = dfs(adj)
        # print(dp, edge, p)
        ans = []
        for i, j in edge:
            mn = min(dp[i], dp[j])
            # print(i, j, mn*(n-mn))
            ans.append(mn*(n-mn))
        ans.sort()
        # print(ans)
        sol = [1]*(n-1)
        st = 0
        pos = 0
        if m<n-1:
            pos = n-1-m
        while st<m and pos<n-1:
            sol[pos] = (sol[pos]*p[st])%md
            st += 1
            pos += 1
        while st<m:
            sol[-1] = (sol[-1]*p[st])%md
            st += 1
        sol1 = 0
        for i in range(n-1):
            sol1 = (sol1 + (ans[i]*sol[i])%md)%md
        # print(sol)
        print(sol1)




main()

"""
7
1 1
6 7 3 4 5 1 3

7
1 2
1 2 6 4 5 3 4
"""





    




                

        

 