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 MOD = 1e9 + 7;
int add(int x, int y) {
  x += y;
  while (x >= MOD) x -= MOD;
  while (x < 0) x += MOD;
  return x;
}
int mul(int x, int y) { return (x * 1ll * y) % MOD; }
int dfs(vector<vector<int>> &g, vector<long long> &ct, int last, int cur,
        int n) {
  if (g[cur].size() == 1 && g[cur][0] == last) return 1;
  int subtree = 0;
  for (int e : g[cur]) {
    if (e == last) continue;
    int nd = dfs(g, ct, cur, e, n);
    subtree += nd;
    ct.push_back(nd * 1ll * (n - nd));
  }
  return subtree + 1;
}
int solve() {
  int n, u, v, m, res = 0;
  cin >> n;
  vector<vector<int>> g(n + 1);
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cin >> m;
  vector<int> p(m), val(n - 1, 1);
  vector<long long> ct;
  for (int i = 0; i < m; i++) cin >> p[i];
  sort(p.rbegin(), p.rend());
  int idx = 0;
  if (m > n - 1) {
    int g = m - (n - 1);
    idx = g;
    for (int i = 0; i < g; i++) p[g] = mul(p[g], p[i]);
  }
  for (int i = idx, j = 0; i < m; i++, j++) val[j] = p[i];
  dfs(g, ct, -1, 1, n);
  sort(ct.rbegin(), ct.rend());
  for (int i = 0; i < n - 1; i++) res = add(res, mul(val[i], ct[i] % MOD));
  return res;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) cout << solve() << "\n";
  return 0;
}
