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;
constexpr int MOD = 1000000007;
template <class T>
void print(T first, string en = " ") {
  for (auto it = first.begin(); it != first.end(); it = next(it)) {
    cout << *it << en[it == prev(first.end())];
  }
  cout << "\n";
}
vector<int> adj[100005];
bool visited[100005];
long long sub[100005];
vector<long long> mul;
int N;
int dfs(int v) {
  visited[v] = true;
  sub[v] = 1;
  for (int u : adj[v]) {
    if (!visited[u]) sub[v] += dfs(u);
  }
  if (sub[v] * (N - sub[v]) != 0) mul.push_back(sub[v] * (N - sub[v]));
  return sub[v];
}
inline int mulmd(long long a, long long b) {
  long long ret = (a * b) % MOD;
  return (int)ret;
}
inline int addmd(long long a, long long b) {
  long long ret = (a + b) % MOD;
  return (int)ret;
}
void solve() {
  cin >> N;
  for (int i = 1; i <= N; i++) {
    adj[i].clear();
  }
  memset(visited, false, sizeof visited);
  memset(sub, 0, sizeof sub);
  mul.clear();
  for (int i = 0; i < N - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int M;
  cin >> M;
  vector<int> fac(M);
  for (int i = 0; i < M; i++) cin >> fac[i];
  sort(fac.rbegin(), fac.rend());
  dfs(1);
  sort(mul.rbegin(), mul.rend());
  long long ans = 0;
  if (mul.size() < fac.size()) {
    int diff = fac.size() - mul.size();
    vector<int> modFac(mul.size(), 1);
    modFac[0] = 1;
    for (int i = 0; i <= diff; i++) {
      modFac[0] = mulmd(modFac[0], fac[i]);
    }
    ans = addmd(ans, mulmd(modFac[0], mul[0]));
    for (int i = diff + 1; i < fac.size(); i++) {
      modFac[i - diff] = mulmd(mul[i - diff], fac[i]);
      ans = addmd(ans, modFac[i - diff]);
    }
  } else {
    for (int i = 0; i < N - 1; i++) {
      if (i >= fac.size()) {
        ans = addmd(ans, mul[i]);
      } else {
        ans = addmd(ans, mulmd(fac[i], mul[i]));
      }
    }
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int T;
  T = 1;
  cin >> T;
  while (T--) {
    solve();
  }
}
