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 int binary_expo(long long int base, long long int pw) {
  long long int res = 1;
  while (pw) {
    if (pw % 2) {
      res *= base;
      res %= 1000000007;
      pw -= 1;
    } else {
      base *= base;
      base %= 1000000007;
      pw /= 2;
    }
  }
  return res;
}
long long int mod_inv(long long int k) {
  return binary_expo(k, 1000000007 - 2);
}
int t, n;
long long dfs_util(int node, vector<vector<int>>& graph,
                   vector<long long>& weights, vector<int>& vis) {
  if (vis[node]) {
    return 0ll;
  }
  vis[node] = 1;
  long long sm = 0;
  for (int j = 0; j < graph[node].size(); j++) {
    if (vis[graph[node][j]]) continue;
    sm += dfs_util(graph[node][j], graph, weights, vis);
  }
  long long val = (sm + 1) * (n - (sm + 1));
  if (val > 0) weights.push_back(val);
  return (sm + 1);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    vector<vector<int>> graph(n + 1);
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      graph[u].push_back(v);
      graph[v].push_back(u);
    }
    int m;
    cin >> m;
    vector<int> facs(m);
    for (int i = 0; i < m; i++) cin >> facs[i];
    sort(facs.begin(), facs.end());
    vector<long long> weights;
    vector<int> vis(n + 1, 0);
    dfs_util(1, graph, weights, vis);
    sort(weights.begin(), weights.end());
    for (int i = 0; i < weights.size(); i++) {
      weights[i] = weights[i] % 1000000007;
    }
    long long ans = 0;
    if (weights.size() < facs.size()) {
      for (int i = 0; i < weights.size() - 1; i++) {
        ans += ((facs[i] * weights[i]) % 1000000007);
        ans = ans % 1000000007;
      }
      long long mul = 1;
      for (int i = weights.size() - 1; i < facs.size(); i++) {
        mul *= (facs[i]);
        mul = mul % 1000000007;
      }
      ans += ((mul * weights[weights.size() - 1]) % 1000000007);
      ans = ans % 1000000007;
      cout << ans << '\n';
    } else {
      int ind = facs.size() - 1;
      int ind1 = weights.size() - 1;
      while (ind >= 0) {
        ans += ((weights[ind1] * facs[ind]) % 1000000007);
        ans = ans % 1000000007;
        ind1--;
        ind--;
      }
      for (int i = ind1; i >= 0; i--) {
        ans += weights[i];
        ans = ans % 1000000007;
      }
      cout << ans << '\n';
    }
  }
  return 0;
}
