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;
int dx[] = {-1, 0, 1, 0, -1, -1, 1, 1};
int dy[] = {0, 1, 0, -1, -1, 1, -1, 1};
const int mod = 1e9 + 7;
long long n;
vector<vector<int>> arr;
vector<long long> p;
long long dfs(int now, int anc) {
  long long num = 0;
  for (int i = 0; i < arr[now].size(); i++) {
    int next = arr[now][i];
    if (next == anc) continue;
    long long x = dfs(next, now);
    p.push_back(x * (n - x));
    num += x;
  }
  return num + 1;
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t, a, b, mm;
  cin >> t;
  while (t--) {
    long long ans = 0;
    cin >> n;
    arr = vector<vector<int>>(n);
    for (long long i = 1; i < n; i++) {
      cin >> a >> b;
      a--;
      b--;
      arr[a].push_back(b);
      arr[b].push_back(a);
    }
    cin >> mm;
    vector<long long> k(mm);
    for (int i = 0; i < mm; i++) {
      cin >> k[i];
    }
    dfs(0, -1);
    if (mm < n) {
      while (k.size() < p.size()) k.push_back(1);
      sort(k.begin(), k.end());
      sort(p.begin(), p.end());
      for (long long i = 0; i < p.size(); i++) {
        ans = (ans + (k[i] * p[i]) % mod) % mod;
      }
    } else {
      sort(k.begin(), k.end());
      sort(p.begin(), p.end());
      for (int i = p.size(); i < mm; i++) {
        k[n - 2] = (k[n - 2] * k[i]) % mod;
      }
      for (long long i = 0; i < p.size(); i++) {
        ans = (ans + (k[i] * p[i]) % mod) % mod;
      }
    }
    cout << ans << '\n';
    arr.clear();
    p.clear();
  }
  return 0;
}
