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 M = 1e9 + 7;
const int N = 1e5 + 5;
long long int n, m;
vector<long long int> adj[N];
long long int a[N];
long long int cnt[N];
void dfs(long long int ch, long long int pa) {
  cnt[ch] = 1;
  for (auto f : adj[ch]) {
    if (pa == f) continue;
    dfs(f, ch);
    cnt[ch] += cnt[f];
  }
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    for (long long int i = 1; i <= n; ++i) adj[i].clear();
    cin >> n;
    vector<pair<long long int, long long int>> vi;
    for (long long int i = 0; i < n - 1; ++i) {
      long long int u, v;
      cin >> u >> v;
      vi.push_back({u, v});
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    cin >> m;
    for (long long int i = 0; i < m; ++i) cin >> a[i];
    sort(a, a + m);
    if (m > n - 1) {
      long long int j = n - 1;
      while (j < m) {
        a[n - 2] = (a[n - 2] * a[j]) % M;
        ++j;
      }
    }
    dfs(1, 0);
    priority_queue<long long int> q;
    for (auto g : vi) {
      long long int c = min(cnt[g.first], cnt[g.second]);
      q.push(c * (n - c));
    }
    long long int ans = 0;
    long long int k = min(n - 2, m - 1);
    while (!q.empty() || k >= 0) {
      if (!q.empty() && k >= 0) {
        long long int f = (q.top()) % M;
        long long int r = a[k] % M;
        q.pop();
        --k;
        r = (r * f) % M;
        ans += r;
        ans %= M;
      } else if (!q.empty()) {
        while (!q.empty()) {
          ans += q.top();
          ans %= M;
          q.pop();
        }
      }
    }
    cout << ans << "\n";
  }
}
