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<long long int> ans;
vector<vector<int> > g(100005);
long long int n;
int visited[100005];
long long int counts[100005];
void dfs(int v, int p) {
  visited[v] = 1;
  counts[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int u = g[v][i];
    if (visited[u] == -1) {
      dfs(u, v);
      counts[v] += counts[u];
      ans.push_back((counts[u] * (n - counts[u])));
    }
  }
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    memset(visited, -1, sizeof(visited));
    for (int i = 1; i <= n; i++) g[i].clear();
    int x, y;
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    memset(counts, 0, sizeof(counts));
    ans.clear();
    dfs(1, -1);
    sort(ans.rbegin(), ans.rend());
    int m;
    cin >> m;
    vector<long long int> k(m);
    for (int i = 0; i < m; i++) cin >> k[i];
    while (k.size() < n - 1) k.push_back(1);
    sort(k.begin(), k.end());
    vector<long long int> k_2;
    int i = k.size() - 1;
    long long int p = k[i--];
    while (i >= 0) {
      if (i >= n - 2)
        p = (p * k[i--]) % 1000000007;
      else if (k_2.size() == 0)
        k_2.push_back(p);
      else
        k_2.push_back(k[i--]);
    }
    if (k_2.size() == 0) k_2.push_back(p);
    p = 0;
    for (int i = 0; i < n - 1; i++)
      p = (p + ((ans[i] % 1000000007) * k_2[i]) % 1000000007) % 1000000007;
    cout << p << endl;
  }
  return 0;
}
