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 os, io

input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
MOD = 10 ** 9 + 7

t = int(input())
for _ in range(t):
    n = int(input())
    adj_lis = []
    adj_cnt = []
    for _ in range(n):
        adj_lis.append([])
        adj_cnt.append(0)
    adj_cnt[0] = 1
    par = [-1] * n
    des_cnt = [1] * n
    leaf_q = []
    edgeNumber = []
    for _ in range(n - 1):
        p, q = map(int, input().split())
        adj_lis[p - 1].append(q - 1)
        adj_lis[q - 1].append(p - 1)
        adj_cnt[p - 1] += 1
        adj_cnt[q - 1] += 1

    treeStack = [(0, -1)]
    while treeStack:
        v, par_v = treeStack.pop()
        par[v] = par_v
        for x in adj_lis[v]:
            if x != par_v:
                treeStack.append((x, v))
        if adj_cnt[v] == 1:
            leaf_q.append(v)

    while leaf_q:
        x = leaf_q.pop()
        if x == 0:
            continue
        edgeNumber.append(des_cnt[x] * (n - des_cnt[x]))
        des_cnt[par[x]] += des_cnt[x]
        adj_cnt[par[x]] -= 1
        if adj_cnt[par[x]] == 1:
            leaf_q.append(par[x])
    edgeNumber.sort()
    m = int(input())
    edgeWeight = []
    primeList = list(map(int, input().split()))
    primeList.sort()
    if n - 1 < m:
        for i in range(n - 1):
            edgeWeight.append(primeList[i])
        for i in range(n - 1, m):
            edgeWeight[n - 2] *= primeList[i]
            edgeWeight[n - 2] %= MOD
    else:
        for i in range(n - 1 - m):
            edgeWeight.append(1)
        for i in range(m):
            edgeWeight.append(primeList[i])
    ans = 0
    for i in range(n - 1):
        ans += edgeNumber[i] * edgeWeight[i]
        ans %= MOD
    print(ans)

