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 printArray(vector<T>& v) {
  for (auto& i : v) cout << i << " ";
  cout << "\n";
}
const int MaxN = 2e5 + 5;
const int MOD = 1e9 + 7;
int n;
vector<int> adj[MaxN];
vector<long long> v;
long long dfs(int u, int p) {
  long long cnt = 1;
  for (auto& v1 : adj[u]) {
    if (v1 == p) continue;
    long long temp = dfs(v1, u);
    v.push_back(temp * (n - temp));
    cnt += temp;
  }
  return cnt;
}
void solve_cases() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int m;
  cin >> m;
  vector<long long> fact;
  for (int i = 0; i < m; i++) {
    int x;
    cin >> x;
    fact.push_back(x);
  }
  dfs(1, -1);
  sort(fact.begin(), fact.end(), greater<long long>());
  sort(v.begin(), v.end(), greater<long long>());
  long long f = 1;
  long long ans = 0;
  if (m < n - 1) {
    for (int i = 0; i < m; i++) {
      ans = (ans + fact[i] * v[i] % MOD) % MOD;
    }
    for (int i = m; i < n - 1; i++) {
      ans = (ans + v[i]) % MOD;
    }
  } else {
    long long tp = 0;
    long long f = 1;
    for (int i = 0; i < m - n + 2; i++) {
      f = (f * fact[i]) % MOD;
      tp++;
    }
    ans = (ans + f * v[0] % MOD);
    for (int i = 1, j = m - n + 2; i < n - 1; i++, j++) {
      ans = (ans + v[i] * fact[j] % MOD) % MOD;
    }
  }
  cout << ans << endl;
  for (int i = 0; i <= n; i++) adj[i].clear();
  v.clear();
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(10) << fixed;
  int t = 1;
  cin >> t;
  for (int tt = 1; tt <= t; tt++) {
    solve_cases();
  }
}
