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;
template <typename T>
void input(vector<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.push_back(temp);
}
template <typename T>
void output(vector<T> arr) {
  for (auto x : arr) cout << x << " ";
  cout << endl;
}
template <typename T>
void input_set(set<T> &arr, long long int n) {
  T temp;
  for (long long int i = 0; i < n; i++) cin >> temp, arr.insert(temp);
}
long long int mul(long long int a, long long int b) {
  return (a % 1000000007 * b % 1000000007) % 1000000007;
}
long long int power(long long int a, long long int b) {
  long long int ans = 1;
  while (b > 0) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    ;
    b >>= 1;
  }
  return ans;
}
vector<vector<int> > adj;
vector<long long int> muls;
vector<bool> visited;
long long int n;
int dfs(int src) {
  long long int nodes = 1;
  visited[src] = true;
  for (auto v : adj[src]) {
    if (visited[v]) continue;
    nodes += dfs(v);
  }
  muls.push_back((n - nodes) * (nodes));
  return nodes;
}
void solve() {
  int u, v, m;
  cin >> n;
  adj.clear();
  adj.resize(n);
  visited.clear();
  visited.resize(n, false);
  muls.clear();
  vector<long long int> p;
  for (int i = 0; i < n - 1; i++) {
    cin >> u >> v;
    --u;
    --v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cin >> m;
  input(p, m);
  int sz = p.size();
  while (sz < n - 1) {
    p.push_back(1);
    sz++;
  }
  sort(p.begin(), p.end());
  while (sz > n - 1) {
    int nn = p.size();
    long long int ele = mul(p[sz - 1], p[sz - 2]);
    p.pop_back();
    p.pop_back();
    p.push_back(ele);
    --sz;
  }
  long long int ans = 0;
  dfs(0);
  sort(muls.begin(), muls.end());
  muls.erase(muls.begin());
  for (int i = 0; i < n - 1; i++)
    ans = (ans % 1000000007 + mul(p[i], muls[i]) % 1000000007) % 1000000007;
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int testcases;
  cin >> testcases;
  while (testcases--) {
    solve();
  }
}
