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;
ifstream in("in.in");
ofstream out("out.out");
struct edge {
  long long u, v, id, cf;
  edge() { u = v = id = cf = 0; }
  edge(long long U, long long V, long long ID) {
    u = U;
    v = V;
    id = ID;
  }
};
vector<edge> edges;
vector<long long> g[100005];
long long p[100005], level[100005], sub[100005], n, m, t;
long long dfs(long long node, long long parinte) {
  sub[node] = 1;
  for (auto k : g[node]) {
    if (k == parinte) continue;
    level[k] = level[node] + 1;
    sub[node] += dfs(k, node);
  }
  return sub[node];
}
void initialize() {
  for (int i = 1; i <= n; i++) sub[i] = level[i] = 0;
  for (int i = 1; i <= n; i++) g[i].clear();
  edges.clear();
}
void solve() {
  long long i;
  dfs(1, 0);
  for (auto k = edges.begin(); k != edges.end(); k++) {
    long long u = (*k).u;
    long long v = (*k).v;
    long long id = (*k).id;
    if (level[u] > level[v]) swap(u, v);
    long long downside = sub[v];
    long long upside = n - downside;
    long long cf = (downside * upside);
    k->cf = cf;
  }
  sort(edges.begin(), edges.end(), [](edge a, edge b) { return a.cf < b.cf; });
  long long sol = 0;
  if (m <= n - 1) {
    for (int i = 1; i <= n - m - 1; i++)
      sol = (sol + edges[i - 1].cf) % 1000000007;
    for (i = n - m; i <= n - 1; i++) {
      sol =
          (sol + edges[i - 1].cf * p[i - n + m + 1] % 1000000007) % 1000000007;
    }
    cout << sol << "\n";
  } else {
    long long nr_cnc = m - n + 2;
    long long conc = 1;
    for (i = m - nr_cnc + 1; i <= m; i++) conc = (conc * p[i]) % 1000000007;
    sol = (sol + conc * edges[n - 2].cf % 1000000007) % 1000000007;
    for (i = 1; i <= n - 2; i++) {
      sol = (sol + p[i] * edges[i - 1].cf % 1000000007) % 1000000007;
    }
    cout << sol << "\n";
  }
}
int main() {
  long long i;
  cin >> t;
  while (t--) {
    initialize();
    cin >> n;
    long long u, v;
    for (int i = 1; i <= n - 1; i++) {
      cin >> u >> v;
      edges.push_back({u, v, i});
      g[u].push_back(v);
      g[v].push_back(u);
    }
    cin >> m;
    for (int i = 1; i <= m; i++) cin >> p[i];
    sort(p + 1, p + 1 + m);
    solve();
    initialize();
  }
}
