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 err(istream_iterator<string> it) {}
template <typename T, typename... Args>
void err(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " ; ";
  err(++it, args...);
  cout << endl;
}
long long n;
long long sz[200000];
vector<long long> g[200000];
long long visit[200000];
priority_queue<long long> pq;
void CLR() {
  while (!pq.empty()) pq.pop();
  for (long long i = 1; i <= n; i++) {
    g[i].clear();
    visit[i] = 0;
    sz[i] = 0;
  }
}
void dfs(long long u) {
  for (auto v : g[u]) {
    if (!visit[v]) {
      visit[v] = 1;
      dfs(v);
      sz[u] += sz[v];
      long long left = sz[v];
      long long right = n - left;
      pq.push(left * right);
    }
  }
  sz[u] += 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    CLR();
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    for (long long i = 1; i <= n; i++) {
      if (g[i].size() == 1) {
        visit[i] = 1;
        dfs(i);
        break;
      }
    }
    long long k;
    cin >> k;
    vector<long long> v;
    for (long long i = 1; i <= k; i++) {
      long long x;
      cin >> x;
      v.push_back(x);
    }
    sort(v.begin(), v.end());
    long long ans = 0;
    while (v.size() > pq.size()) {
      long long keep = v[v.size() - 1];
      long long peek = v[v.size() - 2];
      keep = ((keep % 1000000007) * (peek % 1000000007)) % 1000000007;
      v[v.size() - 2] = keep;
      v.pop_back();
    }
    while (!pq.empty()) {
      long long edgmul = pq.top();
      pq.pop();
      long long keep = 1;
      if (!v.empty()) {
        keep = v[v.size() - 1];
        v.pop_back();
      }
      ans = (ans % 1000000007 + (edgmul * keep) % 1000000007) % 1000000007;
    }
    cout << ans << endl;
  }
  return 0;
}
