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;
int dx[] = {0, 1, 0, -1};
int dy[] = {1, 0, -1, 0};
const long double Pi = acos(-1), e = 2.718, ebs = 0.00000001;
const int N = 5e3 + 10, mod = 1e9 + 7, SEED = 1331, B = 40;
vector<vector<long long> > G;
vector<bool> vis;
vector<long long> cnt, freq;
vector<pair<long long, long long> > E;
long long dfs(int u) {
  vis[u] = 1;
  for (auto v : G[u]) {
    if (!vis[v]) cnt[u] += dfs(v);
  }
  return cnt[u];
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int T;
  cin >> T;
  while (T--) {
    int n;
    cin >> n;
    G.assign(n + 1, vector<long long>());
    vis.assign(n + 1, 0);
    cnt.assign(n + 1, 1);
    E.clear();
    freq.clear();
    for (int i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      G[u].push_back(v);
      G[v].push_back(u);
      E.push_back({u, v});
    }
    int m;
    cin >> m;
    vector<long long> v(m);
    for (auto &x : v) cin >> x;
    if (m < n - 1) {
      while (v.size() < n - 1) v.push_back(1);
    }
    sort(v.begin(), v.end());
    if (m > n - 1) {
      long long x = 1;
      for (int i = n - 2; i < m; i++) {
        x *= v[i];
        x %= mod;
      }
      v.erase(v.begin() + n - 2, v.end());
      v.push_back(x);
    }
    dfs(1);
    long long ans = 0;
    for (auto e : E) {
      long long x = min(cnt[e.first], cnt[e.second]);
      freq.push_back(x * (n - x));
    }
    sort(freq.begin(), freq.end());
    for (int i = 0; i < n - 1; i++) {
      ans += freq[i] * v[i];
      ans %= mod;
    }
    cout << ans << '\n';
  }
  return 0;
}
