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) {
  cout << *it << " = " << a << "\n";
  err(++it, args...);
}
const long long MOD = 1e9 + 7, N = 1e5 + 5;
vector<int> g[N];
vector<long long> c(N, 1), mx;
vector<bool> vis(N, false);
int n;
void dfs(int i) {
  vis[i] = true;
  for (auto x : g[i]) {
    if (!vis[x]) {
      dfs(x);
      c[i] += c[x];
    }
  }
  if (i != 1) {
    mx.push_back(c[i] * (n - c[i]));
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 0; i < n - 1; i++) {
      int x, y;
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    int m;
    cin >> m;
    vector<long long> f(m);
    for (long long i = 0; i < m; i++) cin >> f[i];
    ;
    sort(f.begin(), f.end());
    while (((int)(f).size()) > n - 1) {
      int fl = ((int)(f).size());
      f[fl - 2] = (f[fl - 2] * f[fl - 1]) % MOD;
      f.pop_back();
    }
    reverse(f.begin(), f.end());
    dfs(1);
    sort(mx.begin(), mx.end(), greater<long long>());
    long long ans = 0;
    for (long long i = 0; i < n - 1; i++) {
      if (i < m) {
        ans += (mx[i] * f[i]) % MOD;
        ans %= MOD;
      } else {
        ans += (mx[i] * 1) % MOD;
        ans %= MOD;
      }
    }
    cout << ans << "\n";
    for (long long i = 1; i < n + 1; i++) {
      g[i].clear();
      vis[i] = false;
      c[i] = 1;
    }
    mx.clear();
  }
}
