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> fr;
vector<vector<long long>> g;
long long dfs(int x, int n, int p = -1) {
  if (g[x].size() == 1 && p != -1) {
    fr[x] = 1 * (n - 1);
    return 1;
  }
  long long ctr = 0;
  for (int i = 0; i < g[x].size(); ++i) {
    if (g[x][i] != p) {
      ctr += dfs(g[x][i], n, x);
    }
  }
  fr[x] = (ctr + 1) * (n - ctr - 1);
  return ctr + 1;
}
void solve() {
  long long n, m;
  cin >> n;
  fr = vector<long long>(n, 0);
  g = vector<vector<long long>>(n);
  for (int i = 0; i < n - 1; ++i) {
    long long x, y;
    cin >> x >> y;
    g[x - 1].push_back(y - 1);
    g[y - 1].push_back(x - 1);
  }
  dfs(0, n);
  cin >> m;
  vector<long long> fac(m);
  for (int i = 0; i < m; ++i) {
    cin >> fac[i];
  }
  sort(fac.begin(), fac.end(), greater<long long>());
  sort(fr.begin(), fr.end(), greater<long long>());
  long long sum = 0;
  if (m <= n - 1)
    for (int i = 0; i < n; ++i) {
      if (i < m)
        sum = (sum % 1000000007 + (fr[i] * fac[i]) % 1000000007) % 1000000007;
      else
        sum = (sum % 1000000007 + (fr[i] * 1) % 1000000007) % 1000000007;
    }
  else {
    int i = 0;
    long long p = 1;
    for (; i < m - n + 2; ++i) {
      p = (p % 1000000007 * (fac[i] % 1000000007)) % 1000000007;
    }
    sum = (fr[0] * p) % 1000000007;
    for (int j = 1; j < n; ++j) {
      sum = (sum % 1000000007 + (fr[j] * fac[i]) % 1000000007) % 1000000007;
      i++;
    }
  }
  cout << sum << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
