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;
const int mod = (int)1e9 + 7;
void solve() {
  int n, m;
  scanf("%d", &n);
  vector<vector<int>> G(n, vector<int>());
  vector<long long> f;
  vector<long long> p;
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    G[x].push_back(y);
    G[y].push_back(x);
  }
  function<long long(int, int)> dfs = [&](int u, int pre) {
    long long d = 1;
    for (int& v : G[u]) {
      if (v == pre) continue;
      d += dfs(v, u);
    }
    f.push_back(d * (n - d));
    return d;
  };
  scanf("%d", &m);
  for (int i = 0; i < m; i++) {
    long long x;
    scanf("%lld", &x);
    p.push_back(x);
  }
  dfs(0, -1);
  sort(f.begin(), f.end());
  sort(p.begin(), p.end());
  if (m <= n - 1) {
    for (int i = m; i < n - 1; i++) {
      p.push_back(1LL);
    }
    sort(p.begin(), p.end());
  } else {
    for (int i = n - 1; i < m; i++) {
      p[n - 2] = p[n - 2] * p[i] % mod;
    }
  }
  long long res = 0;
  for (int i = 1; i < (int)f.size(); i++) {
    res = (res + p[i - 1] * f[i] % mod) % mod;
  }
  cout << res << endl;
}
int main() {
  int tt;
  cin >> tt;
  while (tt--) {
    solve();
  }
  return 0;
}
