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 long long MOD = 1000 * 1000 * 1000 + 7;
void dfs(int v, vector<vector<int>> &gr, vector<int> &c, vector<long long> &cnt,
         int p, int n) {
  c[v] = 1;
  for (auto it : gr[v]) {
    if (it != p) {
      dfs(it, gr, c, cnt, v, n);
      c[v] += c[it];
    }
  }
  for (auto it : gr[v]) {
    if (it != p) {
      cnt.push_back(1LL * (n - c[it]) * c[it]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  for (int i = 0; i < t; i++) {
    int n;
    cin >> n;
    vector<vector<int>> gr;
    vector<int> c;
    c.resize(n + 5);
    gr.resize(n + 5);
    for (int j = 0; j < n - 1; j++) {
      int u, v;
      cin >> u >> v;
      gr[u].push_back(v);
      gr[v].push_back(u);
    }
    vector<long long> cnt;
    dfs(1, gr, c, cnt, -1, n);
    sort(cnt.begin(), cnt.end());
    int m;
    cin >> m;
    vector<long long> mmm;
    for (int j = 0; j < m; j++) {
      int a;
      cin >> a;
      mmm.push_back(a);
    }
    while (mmm.size() < cnt.size()) {
      mmm.push_back(1);
    }
    sort(mmm.begin(), mmm.end());
    while (mmm.size() > cnt.size()) {
      long long a = mmm.back();
      mmm.pop_back();
      long long b = mmm.back();
      mmm.pop_back();
      mmm.push_back(a * b % MOD);
    }
    long long res = 0;
    for (int j = 0; j < cnt.size(); j++) {
      res = res + 1LL * cnt[j] * mmm[j];
      res %= MOD;
    }
    cout << res << '\n';
  }
  return 0;
}
