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 n, m;
vector<long long> g[100005];
vector<bool> vis(100005, false);
vector<long long> cnt(100005);
vector<long long> p;
vector<long long> ans;
void dfs(int x) {
  vis[x] = true;
  cnt[x] = 1;
  for (auto i : g[x]) {
    if (!vis[i]) {
      dfs(i);
      ans.push_back(cnt[i] * (n - cnt[i]));
      cnt[x] += cnt[i];
    }
  }
}
int main() {
  long long q = 1, i, j, test, k;
  cin >> q;
  for (test = 1; test < (q + 1); test++) {
    long long x, y;
    cin >> n;
    fill(vis.begin(), vis.end(), false);
    for (i = 0; i < (100005); i++) {
      g[i].clear();
    }
    p.clear();
    ans.clear();
    for (i = 0; i < (n - 1); i++) {
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    cin >> m;
    for (i = 0; i < (m); i++) {
      cin >> k;
      p.push_back(k);
    }
    sort(p.begin(), p.end(), greater<long long>());
    if (m <= n - 1) {
      for (i = m; i < (n - 1); i++) p.push_back(1);
    } else {
      while (p.size() > n - 1) {
        p[1] = (p[1] * p[0]) % 1000000007;
        p.erase(p.begin());
      }
    }
    dfs(1);
    sort(ans.begin(), ans.end(), greater<long long>());
    x = 0;
    for (i = 0; i < (n - 1); i++) {
      x = x + ((ans[i] % 1000000007) * (p[i] % 1000000007)) % 1000000007;
      x %= 1000000007;
    }
    cout << x << '\n';
  }
}
