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> v[100005];
long long int n;
vector<long long int> ed;
long long int dfs(long long int i, long long int p) {
  long long int sum = 1;
  for (long long int j = 0; j < v[i].size(); ++j) {
    long long int node = v[i][j];
    if (node == p) continue;
    long long int tot = dfs(node, i);
    sum += tot;
    ed.push_back(tot * (n - tot));
  }
  return sum;
}
void solve() {
  cin >> n;
  for (long long int i = 1; i < n + 1; ++i) v[i].clear();
  for (long long int i = 0; i < n - 1; ++i) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  long long int m;
  cin >> m;
  vector<long long int> prime;
  for (long long int i = 0; i < m; ++i) {
    long long int x;
    cin >> x;
    prime.push_back(x);
  }
  ed.clear();
  long long int sum = dfs(1, 0);
  sort(prime.begin(), prime.end());
  sort(ed.begin(), ed.end());
  if (m > (n - 1)) {
    vector<long long int> temp;
    for (long long int i = 0; i < n - 2; ++i) {
      temp.push_back(prime[i]);
    }
    long long int pw = 1;
    for (long long int i = n - 2; i < m; ++i) pw = (pw * prime[i]) % 1000000007;
    temp.push_back(pw);
    prime = temp;
  } else if (m < n - 1) {
    long long int diff = n - 1 - m;
    for (long long int i = 0; i < diff; ++i) prime.push_back(1);
    sort(prime.begin(), prime.end());
  }
  long long int ans = 0;
  for (long long int i = 0; i < n - 1; ++i) {
    ed[i] = (ed[i]) % 1000000007;
    long long int temp = (prime[i] * (ed[i])) % 1000000007;
    ans = (ans + temp) % 1000000007;
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return (0);
}
