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;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const long long MOD = 1e9 + 7;
vector<vector<int>> link(100000);
vector<long long> e;
long long n;
long long dfs(int now, int anc) {
  long long x = 0;
  for (int i = 0; i < link[now].size(); i++) {
    int next = link[now][i];
    if (next == anc) continue;
    long long y = dfs(next, now);
    e.push_back(y * (n - y));
    x += y;
  }
  return x + 1;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t, m, a, b;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i < n; i++) {
      cin >> a >> b;
      a--;
      b--;
      link[a].push_back(b);
      link[b].push_back(a);
    }
    cin >> m;
    vector<long long> p(m);
    for (int i = 0; i < m; i++) cin >> p[i];
    dfs(0, -1);
    if (e.size() > p.size()) {
      while (e.size() > p.size()) p.push_back(1);
    }
    sort(e.begin(), e.end());
    sort(p.begin(), p.end());
    if (e.size() < p.size()) {
      for (int i = e.size(); i < p.size(); i++) {
        p[e.size() - 1] = (p[e.size() - 1] * p[i]) % MOD;
      }
    }
    long long ans = 0;
    for (int i = 0; i < e.size(); i++) {
      ans = (ans + ((e[i] * p[i]) % MOD)) % MOD;
    }
    cout << ans << '\n';
    for (int i = 0; i < n; i++) link[i].clear();
    e.clear();
  }
  return 0;
}
