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;
void solve();
string conc(long long int a, long long int b) {
  return to_string(a) + "|" + to_string(b);
}
long long int rev(long long int x) {
  long long int ans = 0;
  while (x != 0) {
    ans = ans * 10 + (x % 10);
    x /= 10;
  }
  return ans;
}
long long int numdig(long long int x) {
  long long int ans = 0;
  while (x != 0) {
    ans++;
    x /= 10;
  }
  return ans;
}
void printvec(vector<long long int> &v) {
  for (long long int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
void printmat(vector<vector<long long int>> &mat) {
  for (long long int i = 0; i < mat.size(); i++) {
    for (long long int j = 0; j < mat[0].size(); j++) cout << mat[i][j] << " ";
    cout << "\n";
  }
}
void dfs(vector<vector<long long int>> &adj, vector<bool> &vis, long long int u,
         vector<long long int> &sub) {
  sub[u] = 1;
  vis[u] = true;
  for (auto v : adj[u]) {
    if (!vis[v]) {
      dfs(adj, vis, v, sub);
      sub[u] += sub[v];
    }
  }
}
void solve() {
  long long int n;
  cin >> n;
  vector<vector<long long int>> adj(n + 1);
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  long long int m;
  cin >> m;
  vector<long long int> primes(m);
  for (long long int i = 0; i < m; i++) cin >> primes[i];
  vector<long long int> sub(n + 1, 0);
  vector<bool> vis(n + 1, false);
  dfs(adj, vis, 1, sub);
  vector<long long int> mults;
  for (long long int i = 2; i <= n; i++)
    mults.push_back((sub[i]) * (n - sub[i]));
  sort(mults.begin(), mults.end(), greater<long long int>());
  sort(primes.begin(), primes.end(), greater<long long int>());
  if (primes.size() < n - 1) {
    while (primes.size() != (n - 1)) primes.push_back(1);
  }
  long long int idx = 0;
  if (primes.size() > n - 1) {
    while (primes.size() - idx != n - 1) {
      primes[idx + 1] *= primes[idx];
      primes[idx + 1] %= 1000000007;
      idx++;
    }
  }
  long long int ans = 0;
  for (long long int i = 0; i < n - 1; i++) {
    long long int ta = (primes[idx++] % 1000000007) * (mults[i] % 1000000007);
    ta %= 1000000007;
    ans += ta;
    ans %= 1000000007;
  }
  cout << ans << "\n";
}
int main(void) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
