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 int64_t mod = 1e9 + 7, N = 1e5 + 5;
vector<int64_t> adj[N];
vector<int64_t> sub(N);
int64_t n;
vector<int64_t> ok;
void dfs(int64_t u, int64_t p) {
  sub[u] = 1;
  for (int64_t &it : adj[u]) {
    if (it == p) {
      continue;
    }
    dfs(it, u);
    sub[u] += sub[it];
  }
}
void dfs1(int64_t u, int64_t p) {
  for (int64_t &it : adj[u]) {
    if (it == p) {
      continue;
    }
    ok.push_back(sub[it] * (n - sub[it]));
    dfs1(it, u);
  }
}
void solve() {
  cin >> n;
  for (int64_t i = 2, x, y; i <= n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  int64_t k;
  cin >> k;
  vector<int64_t> z(k);
  for (int64_t &it : z) {
    cin >> it;
  }
  sort(z.begin(), z.end());
  while (z.size() >= n) {
    int64_t f = z.back();
    z.pop_back();
    z.back() *= f;
    z.back() %= mod;
  }
  dfs(1, 1);
  dfs1(1, 1);
  sort(ok.begin(), ok.end());
  int64_t ans = 0;
  while (ok.size()) {
    if (z.size()) {
      ans = (ans + (ok.back() * z.back()) % mod) % mod;
      z.pop_back();
    } else {
      ans = (ans + ok.back()) % mod;
    }
    ok.pop_back();
  }
  cout << ans << "\n";
  for (int64_t i = 0; i <= n; i++) {
    adj[i].clear();
    sub[i] = 0;
  }
  ok.clear();
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int64_t tq;
  cin >> tq;
  for (; tq; tq--) {
    solve();
  }
}
