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;
vector<long long int> val;
long long int Dfs(vector<long long int> v[], long long int u,
                  long long int past) {
  long long int cnt = 1;
  for (long long int i = 0; i < v[u].size(); i++) {
    if (v[u][i] == past) continue;
    cnt += Dfs(v, v[u][i], u);
  }
  val[u] = cnt;
  return cnt;
}
void UndirGraph(vector<long long int> v[], long long int x, long long int y) {
  v[x].push_back(y);
  v[y].push_back(x);
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n; i++) val.push_back(-1);
    vector<long long int> v[n];
    for (long long int i = 0; i < n - 1; i++) {
      long long int a, b;
      cin >> a >> b;
      a--;
      b--;
      UndirGraph(v, a, b);
    }
    long long int num;
    cin >> num;
    vector<long long int> primes;
    for (long long int i = 0; i < num; i++) {
      long long int hi;
      cin >> hi;
      primes.push_back(hi);
    }
    while (num < n - 1) {
      primes.push_back(1);
      num++;
    }
    sort(primes.begin(), primes.end());
    long long int lolll = Dfs(v, 0, -1);
    vector<long long int> main_val;
    for (long long int i = 0; i < n; i++) {
      if (val[i] == n) continue;
      long long int x;
      x = (n - val[i]) * val[i];
      main_val.push_back(x);
    }
    sort(main_val.begin(), main_val.end());
    long long int lol = 1000000007;
    long long int ans = 0;
    long long int temp_max;
    if (primes.size() >= n - 1)
      temp_max = n - 2;
    else
      temp_max = n - 1;
    for (long long int i = 0; i < temp_max; i++) {
      long long int temp = (main_val[i] % lol) * (primes[i]);
      temp %= lol;
      ans += temp;
    }
    if (primes.size() >= n - 1) {
      long long int k = n - 2;
      long long int temp = 1;
      while (k < primes.size()) {
        temp *= (primes[k]);
        temp %= lol;
        k++;
      }
      ans += (temp * (main_val[n - 2] % lol));
      ans %= lol;
    }
    cout << ans << "\n";
    val.clear();
  }
  return 0;
}
