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 MODULE = 1000000007;
int n, m;
vector<vector<int>> adjacentLists;
vector<int> weights;
vector<pair<int, int>> edges;
vector<long long> weightsEdges, p;
int travel(int cur, int pre) {
  long long temp = 0;
  for (int edge : adjacentLists[cur]) {
    int otherEnd = edges[edge].first + edges[edge].second - cur;
    if (otherEnd == pre) continue;
    temp += travel(otherEnd, cur);
    edges[edge].first = cur;
    edges[edge].second = otherEnd;
  }
  return weights[cur] = temp + 1;
}
void Sol() {
  cin >> n;
  adjacentLists = vector<vector<int>>(n, vector<int>());
  edges = vector<pair<int, int>>(n - 1, pair<int, int>(0, 0));
  weights = vector<int>(n, 0);
  int x, y;
  for (int i = 0; i < n - 1; i++) {
    cin >> edges[i].first >> edges[i].second;
    edges[i].first--;
    edges[i].second--;
    adjacentLists[edges[i].first].push_back(i);
    adjacentLists[edges[i].second].push_back(i);
  }
  travel(0, -1);
  weightsEdges = vector<long long>(n - 1, 0);
  weightsEdges.clear();
  for (pair<int, int>& edge : edges) {
    long long upper = n - weights[edge.second];
    long long lower = weights[edge.second];
    weightsEdges.push_back(upper * lower);
  }
  sort(weightsEdges.begin(), weightsEdges.end());
  cin >> m;
  p = vector<long long>(m, 0);
  for (int i = 0; i < m; i++) cin >> p[i];
  sort(p.begin(), p.end());
  long long res = 0;
  if ((n - 1) < m) {
    for (int i = 0; i < n - 2; i++) {
      weightsEdges[i] %= MODULE;
      res += weightsEdges[i] * p[i];
      res %= MODULE;
    }
    long long temp = weightsEdges[n - 2];
    temp %= MODULE;
    for (int i = n - 2; i < m; i++) {
      temp *= p[i];
      temp %= MODULE;
    }
    res += temp;
    res %= MODULE;
  } else {
    for (int i = 0; i < n - 1 - m; i++) {
      weightsEdges[i] %= MODULE;
      res += weightsEdges[i];
      res %= MODULE;
    }
    for (int i = 0; i < m; i++) {
      weightsEdges[i + n - 1 - m] %= MODULE;
      res += p[i] * weightsEdges[i + n - 1 - m];
      res %= MODULE;
    }
  }
  cout << res;
}
void preprocess() {}
int main() {
  preprocess();
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int num_test = 1;
  cin >> num_test;
  while (num_test--) {
    Sol();
    if (num_test != 0) cout << endl;
  }
}
