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;
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n > 0) {
    if (n % 2 == 1) result = result * x;
    x = x * x;
    n = n / 2;
  }
  return result;
}
void dfs(long long n, vector<vector<long long>>& g, long long par,
         vector<long long>& val, vector<long long>& child) {
  for (auto x : g[n]) {
    if (x == par) continue;
    dfs(x, g, n, val, child);
    child[n] += child[x];
    child[n]++;
  }
  if (par != 0) val.push_back((g.size() - child[n] - 2) * (child[n] + 1));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, ans = 0;
  cin >> n;
  while (n--) {
    cin >> a;
    vector<vector<long long>> V(a + 1, vector<long long>());
    long long B[a + 1][2];
    for (b = 0; b < a - 1; b++) {
      cin >> e >> d;
      V[e].push_back(d);
      V[d].push_back(e);
      B[b][0] = e;
      B[b][1] = d;
    }
    cin >> c;
    vector<long long> A;
    for (d = 0; d < c; d++) {
      cin >> f;
      A.push_back(f);
    }
    while (A.size() < (a - 1)) {
      A.push_back(1);
    }
    sort(A.begin(), A.end());
    vector<long long> D;
    vector<long long> vis(a + 1, 0);
    dfs(1, V, 0, D, vis);
    h = A.size() - 1;
    sort(D.begin(), D.end());
    ans = 0;
    while (h >= D.size()) {
      A[h - 1] = (A[h] * A[h - 1]) % 1000000007;
      h--;
    }
    for (k = D.size() - 1; k >= 0; k--) {
      ans = (ans + (D[k] * A[h]) % 1000000007) % 1000000007;
      h--;
    }
    cout << (ans % 1000000007) << '\n';
  }
  return 0;
}
