#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long MX = 1e5 + 100;
const long long INF = 1e9;
int n, vis[MX];
vector<int> g[MX];
vector<long long> e;
long long dfs(int u) {
  vis[u] = 1;
  long long c = 0, c1;
  for (int v : g[u]) {
    if (!vis[v]) {
      c1 = dfs(v);
      e.push_back(c1 * (n - c1));
      c += c1;
    }
  }
  return c + 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, m, u, v;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i < n + 1; ++i) g[i].clear();
    for (int i = 1; i < n; ++i) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    memset(vis, 0, sizeof(vis));
    e.clear();
    dfs(1);
    cin >> m;
    vector<long long> p(m);
    for (int i = 0; i < m; ++i) cin >> p[i];
    for (int i = m; i < n - 1; ++i) p.push_back(1);
    sort(e.begin(), e.end());
    sort(p.begin(), p.end());
    long long prev = 1;
    m = p.size();
    for (int i = m - 1; i >= n - 1; --i) prev = (prev * p[i]) % MOD;
    long long ans = 0, ans1;
    for (int i = n - 2; i >= 0; --i) {
      e[i] = e[i] % MOD;
      if (prev != 1) p[i] = (p[i] * prev) % MOD;
      ans1 = (e[i] * p[i]) % MOD;
      ans = (ans + ans1) % MOD;
      prev = 1;
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Please formulate a cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1e9 + 7;
const long long MX = 1e5 + 100;
const long long INF = 1e9;
int n, vis[MX];
vector<int> g[MX];
vector<long long> e;
long long dfs(int u) {
  vis[u] = 1;
  long long c = 0, c1;
  for (int v : g[u]) {
    if (!vis[v]) {
      c1 = dfs(v);
      e.push_back(c1 * (n - c1));
      c += c1;
    }
  }
  return c + 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t, m, u, v;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i < n + 1; ++i) g[i].clear();
    for (int i = 1; i < n; ++i) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    memset(vis, 0, sizeof(vis));
    e.clear();
    dfs(1);
    cin >> m;
    vector<long long> p(m);
    for (int i = 0; i < m; ++i) cin >> p[i];
    for (int i = m; i < n - 1; ++i) p.push_back(1);
    sort(e.begin(), e.end());
    sort(p.begin(), p.end());
    long long prev = 1;
    m = p.size();
    for (int i = m - 1; i >= n - 1; --i) prev = (prev * p[i]) % MOD;
    long long ans = 0, ans1;
    for (int i = n - 2; i >= 0; --i) {
      e[i] = e[i] % MOD;
      if (prev != 1) p[i] = (p[i] * prev) % MOD;
      ans1 = (e[i] * p[i]) % MOD;
      ans = (ans + ans1) % MOD;
      prev = 1;
    }
    cout << ans << '\n';
  }
  return 0;
}

```