#include <bits/stdc++.h>
using namespace std;
const int M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
vector<long long> val;
vector<vector<long long>> adj;
long long n;
long long dfs(long long u, long long p) {
  long long ans = 1;
  for (auto v : adj[u]) {
    if (v != p) {
      long long nt = dfs(v, u);
      val.push_back(nt * (n - nt));
      ans += nt;
    }
  }
  return ans;
}
void solve() {
  cin >> n;
  val.clear();
  adj = vector<vector<long long>>(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0LL, -1LL);
  sort(val.begin(), val.end(), greater<long long>());
  int m;
  cin >> m;
  vector<long long> fact(m);
  for (int i = 0; i < m; i++) cin >> fact[i];
  for (int i = m; i < n - 1; i++) fact.push_back(1LL);
  sort(fact.begin(), fact.end(), greater<long long>());
  long long ans = 0;
  if ((long long)fact.size() == n - 1) {
    for (int i = 0; i < n - 1; i++) {
      ans = add(ans, (long long)(mul(fact[i], val[i])));
    }
  } else {
    long long z = fact.size() - (n - 1);
    int x = 0;
    for (long long i = 0; i < (long long)fact.size(); i++) {
      if (i < (z))
        fact[i + 1] = mul(fact[i], fact[i + 1]);
      else
        ans = add(ans, (long long)(mul(fact[i], val[x++])));
    }
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 M = 1e9 + 7;
long long mod(long long x) { return ((x % M + M) % M); }
long long add(long long a, long long b) { return mod(mod(a) + mod(b)); }
long long mul(long long a, long long b) { return mod(mod(a) * mod(b)); }
vector<long long> val;
vector<vector<long long>> adj;
long long n;
long long dfs(long long u, long long p) {
  long long ans = 1;
  for (auto v : adj[u]) {
    if (v != p) {
      long long nt = dfs(v, u);
      val.push_back(nt * (n - nt));
      ans += nt;
    }
  }
  return ans;
}
void solve() {
  cin >> n;
  val.clear();
  adj = vector<vector<long long>>(n);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(0LL, -1LL);
  sort(val.begin(), val.end(), greater<long long>());
  int m;
  cin >> m;
  vector<long long> fact(m);
  for (int i = 0; i < m; i++) cin >> fact[i];
  for (int i = m; i < n - 1; i++) fact.push_back(1LL);
  sort(fact.begin(), fact.end(), greater<long long>());
  long long ans = 0;
  if ((long long)fact.size() == n - 1) {
    for (int i = 0; i < n - 1; i++) {
      ans = add(ans, (long long)(mul(fact[i], val[i])));
    }
  } else {
    long long z = fact.size() - (n - 1);
    int x = 0;
    for (long long i = 0; i < (long long)fact.size(); i++) {
      if (i < (z))
        fact[i + 1] = mul(fact[i], fact[i + 1]);
      else
        ans = add(ans, (long long)(mul(fact[i], val[x++])));
    }
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << fixed;
  cout << setprecision(10);
  int t = 1;
  cin >> t;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}

```