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;
double DEG_to_RAD(double d) { return d * acos(-1.0) / 180.0; }
double RAD_to_DEG(double r) { return r * 180.0 / acos(-1.0); }
long long M = 1e9 + 7, n;
vector<long long> v, d;
vector<vector<long long> > g;
priority_queue<long long> p;
long long dfs(long long t) {
  if (!v[t]) {
    v[t] = 1;
    long long sum = 1;
    for (auto it : g[t]) {
      if (!v[it]) {
        sum += dfs(it);
      }
    }
    return d[t] = sum;
  }
  return d[t];
}
void dfs2(long long t) {
  if (!v[t]) {
    v[t] = 1;
    long long sum = 1;
    for (auto it : g[t]) {
      if (!v[it]) {
        p.push(d[it] * (n - d[it]) * 2);
        dfs2(it);
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long x, y;
    cin >> n;
    g.assign(n, vector<long long>(0));
    for (int i = 0; i < n - 1; i++) {
      cin >> x >> y;
      g[x - 1].push_back(y - 1);
      g[y - 1].push_back(x - 1);
    }
    long long m;
    cin >> m;
    vector<long long> c;
    for (int i = 0; i < m; i++) {
      cin >> x;
      c.push_back(x);
    }
    sort(c.begin(), c.end());
    while (c.size() > n - 1) {
      long long aux = c.back();
      c.pop_back();
      c.back() *= aux;
      c.back() %= M;
    }
    while (c.size() < n - 1) {
      c.push_back(1);
    }
    sort(c.rbegin(), c.rend());
    v.assign(n, 0);
    d.assign(n, 0);
    dfs(0);
    v.assign(n, 0);
    dfs2(0);
    long long ans = 0, in = 0;
    while (!p.empty()) {
      long long t = p.top();
      ans += (t / 2 * c[in]) % M;
      ans %= M;
      in++;
      p.pop();
    }
    cout << ans % M << "\n";
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}
