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 mod = 1e9 + 7;
int n;
vector<int> g[100005];
vector<long long> ev;
int sz[100005];
void dfs(int u, int par) {
  sz[u] = 1;
  for (auto v : g[u]) {
    if (v != par) {
      dfs(v, u);
      sz[u] += sz[v];
    }
  }
  if (par) ev.push_back((1ll * sz[u] * (n - sz[u])));
}
void test_case() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    g[i].clear();
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  ev.clear();
  dfs(1, 0);
  sort(ev.begin(), ev.end());
  for (int i = 0; i < n - 1; i++) {
    ev[i] %= mod;
  }
  int m;
  cin >> m;
  vector<int> a(m);
  for (int i = 0; i < m; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  while (a.size() > n - 1) {
    a[a.size() - 2] = (1ll * a[a.size() - 2] * a.back()) % mod;
    a.pop_back();
  }
  int ans = 0;
  for (int i = 0; i < n - 1; i++) {
    if (i < n - 1 - a.size())
      ans = (ans + ev[i]) % mod;
    else
      ans = (ans + 1ll * ev[i] * a[i - (n - 1 - a.size())]) % mod;
  }
  cout << ans << endl;
  return;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    test_case();
  }
  return 0;
}
