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;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
const int MAX = 1e5;
int n, m;
vector<long long int> adj[MAX];
long long int p[MAX], s[MAX], contrib[MAX];
void dfs(int u = 0, int p = -1) {
  s[u] = 1;
  for (int v : adj[u]) {
    if (v == p) {
      continue;
    }
    dfs(v, u);
    s[u] += s[v];
  }
  contrib[u] = s[u] * (n - s[u]);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      adj[i].clear();
    }
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v, --u, --v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    cin >> m;
    for (int i = 0; i < m; i++) {
      cin >> p[i];
    }
    sort(p, p + m);
    dfs();
    sort(contrib + 1, contrib + n);
    long long int ans = 0;
    for (int i = n - 1; i > 0; i--) {
      long long int pp = 1;
      if (i > n - 1 - m) {
        pp = p[i - 1 - max(n - 1 - m, 0)];
      }
      if (i == n - 1) {
        for (int j = n - 1; j < m; j++) {
          pp = pp * p[j] % 1000000007;
        }
      }
      ans += pp * contrib[i] % 1000000007;
    }
    cout << ans % 1000000007 << "\n";
  }
  return 0;
}
