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;
vector<int> adj[100000];
vector<long long> nvs, pr;
long long n, m;
bool vis[100000];
long long dfs(int node, int child) {
  long long ans = 1, rv = 0;
  for (int i = 0; i < adj[node].size(); i++) {
    if (adj[node][i] == child) continue;
    long long k = dfs(adj[node][i], node);
    rv += k;
    nvs.push_back((k * (n - k)));
  }
  return rv + 1;
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    nvs.clear();
    pr.clear();
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      adj[u - 1].push_back(v - 1);
      adj[v - 1].push_back(u - 1);
    }
    cin >> m;
    for (int i = 0; i < n; i++)
      if (adj[i].size() == 1) {
        long long a = dfs(i, -1);
        break;
      }
    for (int i = 0; i < m; i++) {
      long long p;
      cin >> p;
      pr.push_back(p);
    }
    sort(pr.begin(), pr.end());
    for (int i = 0; i < n; i++) adj[i].clear();
    while (pr.size() > n - 1) {
      long long t = *pr.rbegin();
      pr.pop_back();
      *pr.rbegin() *= t;
      *pr.rbegin() %= ((long long)1e9 + 7);
    }
    sort(nvs.begin(), nvs.end());
    int j = pr.size() - 1;
    long long ans = 0;
    for (int i = n - 2; i >= 0; i--, j--) {
      ans += ((nvs[i]) * (j < 0 ? 1 : pr[j])) % ((long long)1e9 + 7);
      ans %= ((long long)1e9 + 7);
    }
    cout << ans << endl;
  }
  return 0;
}
