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 md = 1e9 + 7;
vector<long long> ans;
vector<int> cnt, was;
vector<vector<int>> go;
void dfs_cnt(int x) {
  cnt[x] = 1;
  for (auto y : go[x]) {
    if (cnt[y] == 0) {
      dfs_cnt(y);
      cnt[x] += cnt[y];
    }
  }
}
void dfs(int x, int up) {
  was[x] = 1;
  for (auto y : go[x]) {
    if (was[y] == 0) {
      int a1 = cnt[y];
      int a2 = up + cnt[x] - cnt[y];
      ans.push_back((long long)a1 * a2);
      dfs(y, a2);
    }
  }
}
vector<int> make_p(const vector<int>& p, int n) {
  vector<int> p2;
  int m = p.size();
  if (m > n - 1) {
    long long cur = 1;
    for (int i = 0; i < m - (n - 1) + 1; i++) cur = (cur * p[i]) % md;
    p2.push_back(cur);
    for (int i = m - (n - 1) + 1; i < m; i++) p2.push_back(p[i]);
  } else {
    p2 = p;
    while ((int)p2.size() < n - 1) p2.push_back(1);
  }
  assert((int)p2.size() == n - 1);
  return p2;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    cin >> n;
    go.clear();
    go.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int x, y;
      cin >> x >> y;
      x--;
      y--;
      go[x].push_back(y);
      go[y].push_back(x);
    }
    int m;
    cin >> m;
    vector<int> p(m);
    for (int i = 0; i < m; i++) cin >> p[i];
    sort(p.begin(), p.end(), greater<int>());
    vector<int> p2 = make_p(p, n);
    cnt.clear();
    ans.clear();
    was.clear();
    cnt.resize(n);
    was.resize(n);
    dfs_cnt(0);
    dfs(0, 0);
    sort(ans.begin(), ans.end(), greater<long long>());
    long long cur = 0;
    for (int i = 0; i < n - 1; i++) {
      long long x = (ans[i] * p2[i]) % md;
      cur = (cur + x) % md;
    }
    cout << cur << endl;
  }
  return 0;
}
