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 dfs(vector<long long> *G, vector<bool> &visited, vector<long long> &f,
              long long v, long long n) {
  visited[v] = true;
  if (G[v].size() == 1 && visited[G[v][0]]) {
    return 1;
  }
  long long m = 1, p;
  for (int u : G[v]) {
    if (!visited[u]) {
      p = dfs(G, visited, f, u, n);
      f.push_back(p * (n - p));
      m += p;
    }
  }
  return m;
}
int main() {
  long long t, u, v, m, n;
  for (cin >> t; t; t--) {
    cin >> n;
    vector<long long> G[n], f;
    vector<bool> visited(n);
    for (int i = n - 1; i; i--) {
      cin >> u >> v;
      G[--u].push_back(--v);
      G[v].push_back(u);
    }
    dfs(G, visited, f, 0, n);
    sort(f.begin(), f.end());
    cin >> m;
    vector<long long> p(m);
    for (int i = 0; i < m; i++) {
      cin >> p[i];
    }
    for (int i = n - 1 - m; i > 0; i--) {
      p.push_back(1);
    }
    sort(p.begin(), p.end());
    long long answer = 0;
    for (int i = 0; i < n - 1; i++) {
      f[i] *= p[i];
    }
    for (int i = n - 1; i < m; i++) {
      f[n - 2] *= p[i];
      f[n - 2] %= 1000000007;
    }
    for (int i = 0; i < n - 1; i++) {
      answer += f[i];
      answer %= 1000000007;
    }
    cout << answer << endl;
  }
}
