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;
const int N = 1e5 + 5;
clock_t timeStart, timeFinish;
void timeBegin() { timeStart = clock(); }
void timeEnd() { timeFinish = clock(); }
void timeDuration() {
  double time_taken = double(timeFinish - timeStart) / double(CLOCKS_PER_SEC);
  cout << "Time taken by program is : " << fixed << time_taken
       << setprecision(5);
  cout << " sec " << endl;
}
int chi[N];
vector<int> adj[N];
vector<long long int> occ;
int n;
void dfs(int x, int par) {
  for (auto w : adj[x]) {
    if (w != par) {
      dfs(w, x);
      chi[x] += chi[w] + 1;
    }
  }
}
void dfs1(int x, int par) {
  for (auto w : adj[x]) {
    if (w != par) {
      dfs1(w, x);
      occ.push_back((chi[w] + 1) * 1LL * (n - chi[w] - 1));
    }
  }
}
void solve() {
  int m, u, v;
  long long int x;
  occ.clear();
  cin >> n;
  for (int i = 1; i < n + 1; i++) chi[i] = 0, adj[i].clear();
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  vector<long long int> vec, vec1;
  cin >> m;
  for (int i = 0; i < m; i++) {
    cin >> x;
    vec.push_back(x);
  }
  sort(vec.begin(), vec.end());
  reverse(vec.begin(), vec.end());
  while (vec.size() < n - 1) vec.push_back(1LL);
  dfs(1, 0);
  dfs1(1, 0);
  sort(occ.begin(), occ.end());
  reverse(occ.begin(), occ.end());
  int sz = vec.size(), ex = 0;
  int j = 1;
  vec1.push_back(vec[0]);
  while (j < sz) {
    if (sz - j > n - 2) {
      vec1[0] *= vec[j];
      vec1[0] %= 1000000007;
    } else if (sz - j <= n - 2) {
      vec1.push_back(vec[j]);
    }
    j++;
  }
  long long int ans = 0;
  for (int i = 0; i < n - 1; i++) {
    ans += vec1[i] * 1LL * occ[i] % 1000000007;
    ans %= 1000000007;
  }
  cout << ans << '\n';
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
