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;
const long long lINF = 1e18 + 7;
const int INF = 1e9 + 7;
template <class T>
istream& operator>>(istream& s, vector<T>& v) {
  for (auto& i : v) {
    s >> i;
  }
  return s;
}
vector<vector<int> > g;
vector<long long> v;
int n, m;
int dfs(int idx, int anc) {
  int sum = 1;
  for (int to : g[idx]) {
    if (to != anc) sum += dfs(to, idx);
  }
  v.push_back(1ll * sum * (n - sum));
  return sum;
}
void solve(int CASE = 0) {
  cin >> n;
  g = vector<vector<int> >(n + 1);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  cin >> m;
  priority_queue<long long> p;
  for (int i = 0; i < m; i++) {
    int a;
    cin >> a;
    p.push(a);
  }
  v.clear();
  dfs(1, 0);
  sort((v).begin(), (v).end());
  while (p.size() >= n) {
    long long a = p.top() % INF;
    p.pop();
    long long b = p.top() % INF;
    p.pop();
    p.push((a * b) % INF + INF);
  }
  long long ans = 0;
  for (int i = n - 1; i > 0; i--) {
    v[i] %= INF;
    long long t = 1;
    if (!p.empty()) {
      t = p.top() % INF;
      p.pop();
    }
    ans += v[i] * t % INF;
    ans %= INF;
  }
  cout << ans << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int q = 1;
  cin >> q;
  for (int i = 1; i <= q; i++) {
    solve(i);
  }
}
