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;
using ll = long long;
const int N = 1e5 + 5;
ll mod = 1e9 + 7;
vector<int> g[N];
bool vis[N];
ll sz[N];
void dfs(int s) {
  vis[s] = true;
  sz[s] = 1;
  for (auto u : g[s]) {
    if (!vis[u]) {
      dfs(u);
      sz[s] += sz[u];
    }
  }
}
void solve() {
  ll n;
  cin >> n;
  for (int i = 1; i <= n; ++i) g[i].clear();
  fill(sz, sz + n + 1, 0);
  fill(vis, vis + n + 1, false);
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  int m;
  cin >> m;
  vector<ll> p(m);
  for (int i = 0; i < m; ++i) cin >> p[i];
  while (p.size() < n - 1) {
    p.push_back(1);
  }
  dfs(1);
  vector<ll> times;
  for (int i = 1; i <= n; ++i) {
    times.push_back(sz[i] * (n - sz[i]));
  }
  sort(times.rbegin(), times.rend());
  sort(p.rbegin(), p.rend());
  ll val = 1;
  while (p.size() > n - 1) {
    val = val * p[0];
    val %= mod;
    p.erase(p.begin());
  }
  p[0] = p[0] * val;
  p[0] %= mod;
  ll ans = 0;
  for (int i = 0; i < n - 1; ++i) {
    times[i] %= mod;
    p[i] %= mod;
    ans = ans + times[i] * p[i];
    ans %= mod;
  }
  cout << ans << '\n';
  return;
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
