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 t;
long long n, np;
long long visited[100005 + 1];
long long noOfchildren[100005 + 1];
long long par[100005 + 1];
long long dfs(long long i, vector<vector<long long>> &g, long long par[]) {
  visited[i] = 1;
  long long children = 1;
  for (long long j = 0; j < g[i].size(); j++) {
    if (visited[g[i][j]])
      par[i] = g[i][j];
    else
      children += dfs(g[i][j], g, par);
  }
  return noOfchildren[i] = children;
}
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    pair<long long, long long> e[n + 1];
    vector<vector<long long>> g(n + 1, vector<long long>(0, 0));
    for (long long i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      e[i + 1] = {x, y};
      g[x].push_back(y);
      g[y].push_back(x);
    }
    cin >> np;
    vector<long long> Pr(np, 0);
    for (long long i = 0; i < np; i++) cin >> Pr[i];
    sort(Pr.begin(), Pr.end(), greater<long long>());
    memset(par, 0, sizeof(par));
    memset(noOfchildren, 0, sizeof(noOfchildren));
    memset(visited, 0, sizeof(visited));
    dfs(1, g, par);
    vector<long long> ans;
    for (long long i = 1; i <= n - 1; i++) {
      if (par[e[i].first] == e[i].second)
        ans.push_back(noOfchildren[e[i].first] *
                      (n - noOfchildren[e[i].first]));
      else
        ans.push_back(noOfchildren[e[i].second] *
                      (n - noOfchildren[e[i].second]));
    }
    long long sol = 0;
    long long no = 1;
    sort(ans.begin(), ans.end(), greater<long long>());
    if (np < n) {
      for (long long i = 0; i < n - 1; i++)
        if (i < np)
          sol = sol + (Pr[i] * ans[i]) % 1000000007;
        else
          sol = sol + ans[i];
      cout << (sol) % 1000000007 << endl;
    } else {
      for (long long i = 0; i < np - (n - 2); i++)
        no = (no * Pr[i]) % 1000000007;
      sol += (no * ans[0]) % 1000000007;
      for (long long i = np - (n - 2); i < np; i++)
        sol = (sol + Pr[i] * ans[i + 1 - (np - (n - 2))]) % 1000000007;
      cout << (sol) % 1000000007 << endl;
    }
  }
  return 0;
}
