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 power(long long x, unsigned long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long modInverse(long long n, long long p) { return power(n, p - 2, p); }
void dfs(vector<vector<long long> > &adj, long long n, vector<long long> &s,
         long long i, long long x, vector<long long> &p) {
  s[i] = 1;
  for (auto j : adj[i]) {
    if (j != x) {
      p[j] = i;
      dfs(adj, n, s, j, i, p);
      s[i] += s[j];
    }
  }
  return;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long T = 1;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    vector<vector<long long> > adj(n + 1);
    vector<long long> s(n + 1, 0);
    vector<long long> p(n + 1, -1);
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      cin >> a >> b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    long long k;
    cin >> k;
    vector<long long> temp(k);
    for (long long i = 0; i < k; i++) {
      cin >> temp[i];
    }
    sort(temp.begin(), temp.end());
    reverse(temp.begin(), temp.end());
    map<pair<long long, long long>, long long> m;
    dfs(adj, n, s, 1, -1, p);
    vector<long long> v;
    for (long long i = 2; i <= n; i++) {
      v.push_back((n - s[i]) * s[i]);
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    long long i = 0;
    while (k >= n) {
      temp[i + 1] = (temp[i] * temp[i + 1]) % 1000000007;
      k--;
      i++;
    }
    long long ans = 0;
    for (long long j = 0; j < n - 1; j++) {
      if (i < temp.size())
        ans = (ans + (temp[i] * v[j]) % 1000000007) % 1000000007;
      else {
        ans = (ans + v[j]) % 1000000007;
      }
      i++;
    }
    cout << ans << "\n";
  }
  return 0;
}
