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;
long long INF = 1000000000000000005LL;
void solve();
long long dfs(vector<vector<long long> > &g, vector<bool> &vis,
              vector<long long> &ch, vector<long long> &par, long long u) {
  if (vis[u]) {
    return ch[u];
  }
  vis[u] = true;
  if (g[u].size() == 1 && u != 1) {
    ch[u] = 1;
    return 1;
  }
  long long c = 0;
  for (auto x : g[u]) {
    if (not vis[x]) {
      c += dfs(g, vis, ch, par, x);
      par[x] = u;
    }
  }
  ch[u] = 1 + c;
  return ch[u];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  vector<vector<long long> > g(n + 1);
  map<pair<long long, long long>, long long> d;
  long long u, v;
  for (long long i = 0; i < n - 1; i++) {
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
    d[{u, v}] = 0;
  }
  long long m;
  cin >> m;
  vector<long long> p(m);
  for (long long i = 0; i < m; i++) {
    cin >> p[i];
  }
  vector<bool> vis(n + 1, false);
  vector<long long> ch(n + 1);
  vector<long long> par(n + 1);
  dfs(g, vis, ch, par, 1);
  par[1] = -1;
  vector<long long> freq;
  long long fk, sk, cnt;
  for (auto x : d) {
    fk = x.first.first;
    sk = x.first.second;
    if (par[fk] == sk) {
      cnt = ch[fk] * (n - ch[fk]);
    } else {
      cnt = ch[sk] * (n - ch[sk]);
    }
    freq.push_back(cnt);
  }
  long long ans = 0;
  if (m <= n - 1) {
    for (long long i = 0; i < n - 1 - m; i++) {
      p.push_back(1);
    }
    sort(freq.rbegin(), freq.rend());
    sort(p.rbegin(), p.rend());
    for (long long i = 0; i < n - 1; i++) {
      ans = (ans % 1000000007) +
            ((freq[i] % 1000000007) * (p[i] % 1000000007)) % 1000000007;
      ans = ans % 1000000007;
    }
    cout << ans % 1000000007 << '\n';
    return;
  } else {
    long long end = 1;
    sort(freq.begin(), freq.end());
    sort(p.begin(), p.end());
    for (long long i = n - 2; i < m; i++) {
      end = ((end % 1000000007) * (p[i] % 1000000007)) % 1000000007;
    }
    p[n - 2] = end;
    for (long long i = 0; i < n - 1; i++) {
      ans = (ans % 1000000007) +
            ((freq[i] % 1000000007) * (p[i] % 1000000007)) % 1000000007;
      ans = ans % 1000000007;
    }
    cout << ans % 1000000007 << '\n';
  }
  return;
}
