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;
template <class T>
bool ckmin(T& a, const T& b) {
  return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
  return a < b ? a = b, 1 : 0;
}
vector<vector<long long> > adj(2e5);
vector<long long> val;
long long n;
long long dfs(long long curr, long long par) {
  if (adj[curr].size() == 0) {
    return 0;
  }
  long long ans = 1;
  for (int i = 0; i < (long long)adj[curr].size(); i++) {
    if (adj[curr][i] != par) {
      long long nxt = dfs(adj[curr][i], curr);
      val.push_back(nxt * (n - nxt));
      ans += nxt;
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long mod = 1e9 + 7;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < 2e5; i++) {
      adj[i].clear();
    }
    val.clear();
    for (int i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      x--, y--;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    long long m;
    cin >> m;
    vector<long long> prime(m);
    for (int i = 0; i < m; i++) {
      cin >> prime[i];
      prime[i] %= mod;
    }
    dfs(0, -1);
    sort(val.begin(), val.end());
    reverse(val.begin(), val.end());
    sort(prime.begin(), prime.end());
    reverse(prime.begin(), prime.end());
    if (m <= n - 1) {
      long long ans = 0;
      for (int i = 0; i < m; i++) {
        val[i] %= mod;
        ans += (val[i] * prime[i]) % mod;
        ans %= mod;
      }
      for (int i = m; i < n - 1; i++) {
        ans += (val[i]) % mod;
        ans %= mod;
      }
      cout << ans << "\n";
    } else {
      long long ans = 0;
      long long sz = m - (n - 1);
      for (int i = 0; i < sz; i++) {
        prime[i + 1] = (prime[i + 1] * prime[i]) % mod;
        prime[i + 1] %= mod;
      }
      for (int i = (m - 1) - (n - 1) + 1; i < m; i++) {
        ans += (prime[i] * val[i - ((m - 1) - (n - 1) + 1)]) % mod;
        ans %= mod;
      }
      cout << ans << "\n";
    }
  }
  return 0;
}
