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 int64 = long long;
void DBG() { cerr << "]" << '\n'; }
template <class H, class... T>
void DBG(H h, T... t) {
  cerr << h;
  if (sizeof...(t)) cerr << ", ";
  DBG(t...);
}
const int MOD = (int)(1e9) + 7;
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<vector<int> > g(n + 1);
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      g[a].push_back(b);
      g[b].push_back(a);
    }
    int m;
    cin >> m;
    vector<int64> v(max(n - 1, m), 1);
    for (int i = 0; i < m; i++) cin >> v[i];
    sort(v.begin(), v.end());
    for (int i = n - 1; i < (int)v.size(); i++)
      v[n - 2] = v[n - 2] * v[i] % MOD;
    v.erase(v.begin() + n - 1, v.end());
    assert((int)v.size() == n - 1);
    vector<int> sz(n + 1);
    vector<int64> c;
    function<void(int, int)> dfs = [&](int u, int p) {
      sz[u] = 1;
      for (int v : g[u]) {
        if (v != p) {
          dfs(v, u);
          c.push_back(1LL * sz[v] * (n - sz[v]));
          sz[u] += sz[v];
        }
      }
    };
    dfs(1, -1);
    sort(c.begin(), c.end());
    int64 ans = 0;
    for (int i = 0; i < n - 1; i++) {
      int value = v[i] * c[i] % MOD;
      ans = (ans + value) % MOD;
    }
    cout << ans << '\n';
  }
  return 0;
}
