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;
inline void get(vector<long long> &arr, long long n) {
  long long a;
  for (long long i = 0; i < n; i++) {
    cin >> a;
    arr.push_back(a);
  }
}
inline void print(vector<long long> &arr, long long n) {
  for (long long i = 0; i < n; i++) cout << arr[i] << " ";
  cout << "\n";
  ;
}
vector<long long> edges;
vector<vector<long long>> adj(100005, vector<long long>());
vector<bool> vis(100005);
long long nodes;
long long dfs(long long node) {
  vis[node] = true;
  long long sum = 1;
  for (long long i = 0; i < adj[node].size(); i++) {
    if (!vis[adj[node][i]]) sum += dfs(adj[node][i]);
  }
  long long children = ((nodes - sum) * sum);
  edges.push_back(children);
  return (sum);
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> nodes;
    for (long long i = (0); i < (nodes); ++i) vis[i] = false;
    for (long long i = (0); i < (nodes - 1); ++i) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    dfs(1);
    long long m;
    cin >> m;
    vector<long long> factors(m);
    for (long long i = (0); i < (m); ++i) cin >> factors[i];
    sort(factors.begin(), factors.end());
    while (factors.size() > nodes - 1) {
      long long temp = factors.back();
      factors.pop_back();
      factors.back() =
          (factors.back() % 1000000007 * temp % 1000000007) % 1000000007;
    }
    reverse(factors.begin(), factors.end());
    while (factors.size() < nodes - 1) factors.push_back(1);
    sort(edges.begin(), edges.end(), greater<long long>());
    long long ans = 0;
    for (long long i = (0); i < (nodes - 1); ++i) {
      ans =
          (ans % 1000000007 + factors[i] * edges[i] % 1000000007) % 1000000007;
    }
    for (long long i = (0); i < (nodes + 1); ++i) adj[i].clear();
    for (long long i = (0); i < (nodes + 1); ++i) vis[i] = false;
    edges.clear();
    cout << ans << "\n";
    ;
  }
  return 0;
}
