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.
#include <bits/stdc++.h>
using namespace std;
long long int n;
vector<long long int> adj[(long long int)(1e5 + 5)], d;
long long int dfs(long long int c, long long int pre) {
  long long int a = 1;
  for (auto x : adj[c]) {
    if (x == pre) continue;
    long long int b = dfs(x, c);
    d.push_back((n - b) * b);
    a += b;
  }
  return a;
}
void solve() {
  for (long long int i = 0; i <= (long long int)(1e5 + 5); i++) adj[i].clear();
  d.clear();
  cin >> n;
  for (long long int i = 1; i < n; i++) {
    long long int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, -1);
  long long int m;
  cin >> m;
  vector<long long int> pr(max(m, n), 1);
  for (long long int i = 0; i < m; i++) cin >> pr[i];
  sort(d.begin(), d.end());
  reverse(d.begin(), d.end());
  sort(pr.begin(), pr.end());
  reverse(pr.begin(), pr.end());
  long long int ans = 0, cur = 0;
  if (n - 1 < m)
    for (; cur < m - n + 1; cur++)
      pr[cur + 1] = (pr[cur + 1] * pr[cur]) % (long long int)(1e9 + 7);
  for (long long int i = 0; i < n - 1; i++)
    ans = (ans + (pr[i + cur] * d[i] % (long long int)(1e9 + 7))) %
          (long long int)(1e9 + 7);
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
