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;
using P = pair<int, int>;
using pll = pair<long long, long long>;
constexpr int INF = 1e9;
constexpr long long LINF = 1e17;
constexpr int MOD = 1000000007;
constexpr double PI = 3.14159265358979323846;
vector<int> tree[100005];
ll num[100005];
ll dfs(const int& n, int i, int before) {
  ll res = 1;
  for (int j : tree[i]) {
    if (j == before) continue;
    res += dfs(n, j, i);
  }
  num[i] = (res) * (ll)(n - res);
  return res;
}
void solve() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < (int)(n); i++) tree[i] = {}, num[i] = 0;
  for (int i = 0; i < (int)(n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    tree[u].push_back(v);
    tree[v].push_back(u);
  }
  dfs(n, n - 1, -1);
  sort(num, num + n - 1);
  int m;
  scanf("%d", &m);
  vector<ll> p(m);
  for (int i = 0; i < (int)(m); i++) scanf("%lld", &p[i]);
  vector<ll> q(n - 1, 1);
  if (m <= n - 1) {
    for (int i = 0; i < (int)(m); i++) q[i] = p[i];
    sort(q.begin(), q.end());
  } else {
    sort(p.begin(), p.end());
    for (int i = 0; i < (int)(n - 2); i++) q[i] = p[i];
    for (int i = n - 2; i < m; i++) q[n - 2] *= p[i], q[n - 2] %= MOD;
  }
  ll ans = 0;
  for (int i = 0; i < (int)(n - 1); i++) {
    ans = (ans + q[i] * num[i]) % MOD;
  }
  cout << ans << endl;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    solve();
  }
  return 0;
}
