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;
void Main();
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  Main();
  return 0;
}
const int N = 1e5 + 10, MOD = 1e9 + 7;
vector<int> edg[N];
long long cnt[N];
int siz[N], n;
void dfs(int u, int f) {
  siz[u] = 1;
  for (const int& v : edg[u]) {
    if (v != f) {
      dfs(v, u);
      siz[u] += siz[v];
    }
  }
  cnt[u] = 1ll * (n - siz[u]) * siz[u];
}
void Main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) edg[i].clear();
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      edg[u].push_back(v);
      edg[v].push_back(u);
    }
    int m;
    cin >> m;
    vector<int> v(m);
    for (int i = 0; i < m; i++) {
      cin >> v[i];
    }
    while (int(v.size()) < n - 1) {
      v.push_back(1);
    }
    sort(v.begin(), v.end());
    while (int(v.size()) > n - 1) {
      const int a = v.back();
      v.pop_back();
      const int b = v.back();
      v.pop_back();
      v.push_back(1ll * a * b % MOD);
    }
    dfs(1, -1);
    vector<int> id(n);
    iota(id.begin(), id.end(), 1);
    sort(id.begin(), id.end(),
         [&](const int& x, const int& y) { return cnt[x] < cnt[y]; });
    int ans = 0;
    for (int i = 1; i < n; i++) {
      const int j = id[i];
      ans += cnt[j] % MOD * v[i - 1] % MOD;
      if (ans >= MOD) ans -= MOD;
    }
    cout << ans << '\n';
  }
}
