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>
int a;
int size[100010];
std::vector<int> V[100010];
std::vector<long long int> save;
int getSize(int k, int prev) {
  if (size[k]) return size[k];
  int ans = 1;
  for (int i = 0; i < V[k].size(); i++) {
    if (V[k][i] == prev) continue;
    ans += getSize(V[k][i], k);
  }
  return size[k] = ans;
}
void func(int k, int prev) {
  for (int i = 0; i < V[k].size(); i++) {
    if (V[k][i] == prev) continue;
    long long int t = getSize(V[k][i], k);
    save.push_back(t * (long long int)(a - t));
    func(V[k][i], k);
  }
}
int x[100010];
long long int y[100010];
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &a);
    for (int i = 1; i <= a; i++) size[i] = 0;
    for (int i = 1; i <= a; i++) V[i].clear();
    save.clear();
    for (int i = 1; i < a; i++) {
      int b, c;
      scanf("%d%d", &b, &c);
      V[b].push_back(c);
      V[c].push_back(b);
    }
    func(1, 0);
    std::sort(save.begin(), save.end());
    int b;
    scanf("%d", &b);
    for (int i = 1; i <= b; i++) scanf("%d", &x[i]);
    std::sort(x + 1, x + b + 1);
    if (b <= a - 1) {
      for (int i = 1; i <= (a - 1 - b); i++) y[i] = 1;
      for (int i = (a - b); i <= a - 1; i++) y[i] = x[i - (a - b) + 1];
    } else {
      y[a - 1] = 1;
      for (int i = 1; i <= a - 2; i++) y[i] = x[i];
      for (int i = a - 1; i <= b; i++) {
        y[a - 1] *= x[i];
        y[a - 1] %= 1000000007;
      }
    }
    long long int ans = 0;
    for (int i = 0; i < save.size(); i++) save[i] %= 1000000007;
    for (int i = 1; i <= a - 1; i++) {
      long long int S = save[i - 1] * y[i];
      ans += (S % 1000000007);
      ans %= 1000000007;
    }
    printf("%lld\n", ans);
  }
}
