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;
const long long MOD = 1e9 + 7;
vector<vector<long long>> edges;
vector<long long> multi;
long long dfs(long long root, long long parent = -1) {
  long long count = 0;
  for (auto& child : edges[root]) {
    if (child == parent) continue;
    long long temp = dfs(child, root);
    count += temp;
    multi.push_back(temp);
  }
  return count + 1;
}
void solve() {
  long long n;
  cin >> n;
  edges.clear();
  edges.resize(n, vector<long long>(0));
  multi.clear();
  for (int i = 0, _n = (n - 1); i < _n; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    edges[u].push_back(v);
    edges[v].push_back(u);
  }
  dfs(0);
  for (int i = 0, _n = (multi.size()); i < _n; i++) {
    multi[i] *= (n - multi[i]);
  }
  sort((multi).begin(), (multi).end(), greater<long long>());
  for (int i = 0, _n = (n - 1); i < _n; i++) multi[i] %= MOD;
  long long m;
  cin >> m;
  vector<long long> factors(m);
  for (int i = 0, _n = (m); i < _n; i++) cin >> factors[i];
  sort((factors).begin(), (factors).end(), greater<long long>());
  long long ans = 0;
  if (factors.size() <= n - 1) {
    for (int i = 0, _n = (factors.size()); i < _n; i++) {
      ans = (ans % MOD) + (factors[i] * multi[i]) % MOD;
      ans %= MOD;
    }
    for (int i = (factors.size()), _b = (n - 2); i <= _b; i++) {
      ans += (multi[i]) % MOD;
      ans %= MOD;
    }
  } else {
    long long extra = factors.size() - n + 1;
    long long first = 1;
    for (int i = 0, _n = (extra); i < _n; i++) {
      first *= factors[i];
      first %= MOD;
    }
    factors[extra] *= first;
    factors[extra] %= MOD;
    for (int i = 0, _n = (n - 1); i < _n; i++) {
      ans = (ans % MOD + (multi[i] * factors[extra + i]) % MOD) % MOD;
      ans %= MOD;
    }
  }
  cout << (ans + MOD) % MOD << '\n';
  return;
}
int main(int argc, char** argv) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
