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 long long mod = 1e9 + 7;
vector<int> G[100005];
long long a[100005], d[100005], n, m;
vector<long long> nums;
void dfs(int u, int f) {
  d[u] = 1;
  for (int i = 0; i < G[u].size(); ++i) {
    int x = G[u][i];
    if (x == f) continue;
    dfs(x, u);
    d[u] += d[x];
  }
}
void dfs2(int u, int f) {
  for (int i = 0; i < G[u].size(); ++i) {
    int x = G[u][i];
    if (x == f) continue;
    nums.push_back(d[x] * (n - d[x]));
    dfs2(x, u);
  }
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> n;
    nums.clear();
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 1, u, v; i < n; ++i)
      cin >> u >> v, G[u].push_back(v), G[v].push_back(u);
    cin >> m;
    for (int i = 0; i < m; ++i) cin >> a[i];
    sort(a, a + m);
    dfs(1, 0), dfs2(1, 0);
    sort(nums.begin(), nums.end());
    long long ans = 0;
    int j = 0;
    if (m > n - 1) {
      for (int i = n - 1; i < m; ++i) a[n - 2] = a[n - 2] * a[i] % mod;
      for (int i = 0; i < n - 1; ++i)
        ans = (ans + nums[i] % mod * a[i] % mod) % mod;
    } else {
      for (int i = 0; i < n - 1 - m; ++i) ans = (ans + nums[j++] % mod) % mod;
      for (int i = 0; i < m; ++i)
        ans = (ans + nums[j++] % mod * a[i] % mod) % mod;
    }
    cout << ans << endl;
  }
  return 0;
}
