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;
long long const lim = 1e5 + 5;
long long const mod = 1000000007;
vector<long long> adj[lim + 5];
long long vis[lim + 5], n;
vector<long long> count_edge;
long long child[lim + 5];
void all_clear() {
  count_edge.clear();
  for (long long i = 1; i <= lim; i++) {
    adj[i].clear();
    vis[i] = -1;
    child[i] = 0;
  }
}
void dfs(long long s) {
  vis[s] = 1;
  for (auto v : adj[s]) {
    if (vis[v] == -1) {
      dfs(v);
      child[s] += child[v];
      count_edge.push_back(child[v] * (n - child[v]));
    }
  }
  child[s] += 1;
}
void solve() {
  cin >> n;
  all_clear();
  for (int i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  long long m;
  cin >> m;
  vector<long long> fact;
  for (int i = 0; i < m; i++) {
    long long p;
    cin >> p;
    fact.push_back(p);
  }
  dfs(1);
  sort(count_edge.rbegin(), count_edge.rend());
  sort(fact.rbegin(), fact.rend());
  for (int i = 0; i < count_edge.size(); i++) count_edge[i] %= mod;
  long long ans = 0;
  if (fact.size() > count_edge.size()) {
    while (fact.size() > count_edge.size()) {
      long long val = fact[0];
      fact.erase(fact.begin());
      fact[0] = (fact[0] * val) % mod;
    }
    for (int i = 0; i < fact.size(); i++) {
      ans = (ans + (fact[i] * count_edge[i]) % mod) % mod;
    }
  } else {
    for (int i = 0; i < count_edge.size(); i++) {
      if (i >= fact.size()) {
        ans = (ans + count_edge[i]) % mod;
      } else {
        ans = (ans + (fact[i] * count_edge[i]) % mod) % mod;
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int t = 1, no = 0;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
