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 int MAXN = 3e5 + 5;
const int INF = 0x3f3f3f3f;
const long long p = 1e9 + 7;
int n, m, N, a[MAXN];
long long b[MAXN], c[MAXN];
vector<int> G[MAXN];
int dfs(int u, int fa) {
  int res = 1, x;
  for (int& v : G[u])
    if (v != fa) {
      x = dfs(v, u), res += x;
      b[++N] = 1ll * x * (n - x);
    }
  return res;
}
inline void run() {
  cin >> n, N = 0;
  for (int i = (int)(1); i <= (int)(n); i++) G[i].clear();
  int u, v;
  for (int i = (int)(2); i <= (int)(n); i++) {
    cin >> u >> v;
    G[u].emplace_back(v);
    G[v].emplace_back(u);
  }
  dfs(1, 0);
  sort(b + 1, b + N + 1, greater<long long>());
  cin >> m;
  for (int i = (int)(1); i <= (int)(m); i++) cin >> a[i];
  sort(a + 1, a + m + 1, greater<int>());
  long long res = 0;
  if (N > m) {
    for (int i = (int)(1); i <= (int)(m); i++)
      res = (res + a[i] * b[i] % p) % p;
    for (int i = (int)(m + 1); i <= (int)(N); i++) res = (res + b[i]) % p;
  } else {
    for (int i = (int)(N); i >= (int)(1); i--) c[i] = a[m--];
    while (m) c[1] = c[1] * a[m--] % p;
    for (int i = (int)(1); i <= (int)(N); i++)
      res = (res + c[i] * b[i] % p) % p;
  }
  cout << (res) << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  int T;
  for (cin >> T; T--;) run();
  return 0;
}
