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;
using ll = long long;
using ld = long double;
int INF = INT_MAX;
const int N = 100 * 1000 + 5;
const long long mod = 1000000007;
int n, u, v, m;
vector<vector<int>> gp;
vector<int> p;
vector<long long> mult;
int dfs(int v, int prev = -1) {
  int cnt = 0;
  for (int u : gp[v]) {
    if (u == prev) continue;
    int tmp = dfs(u, v);
    cnt += tmp;
    mult.push_back(tmp * 1ll * (n - tmp));
  }
  return cnt + 1;
}
void solve_for_test(int test_id) {
  cin >> n;
  gp.resize(n);
  p.clear();
  mult.clear();
  gp.assign(n, vector<int>());
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    --u;
    --v;
    gp[u].push_back(v);
    gp[v].push_back(u);
  }
  cin >> m;
  p.resize(m);
  for (int i = 0; i < m; i++) {
    cin >> p[i];
  }
  while (p.size() < n - 1) {
    p.push_back(1);
  }
  sort(p.begin(), p.end());
  while (p.size() > n - 1) {
    p[p.size() - 2] = (p[p.size() - 2] * 1ll * p[p.size() - 1]) % mod;
    p.pop_back();
  }
  for (int i = 0; i < n; i++) {
    if (gp[i].size() == 1) {
      dfs(i);
      break;
    }
  }
  sort(mult.begin(), mult.end());
  long long ans = 0;
  for (int i = 0; i < n - 1; i++) {
    (ans += p[i] * 1ll * mult[i]) %= mod;
  }
  cout << ans << endl;
}
int main() {
  int test_cases = 1;
  cin >> test_cases;
  for (int test_id = 0; test_id < test_cases; ++test_id) {
    solve_for_test(test_id);
  }
  return 0;
}
