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 mi(long long n, long long m) {
  long long pw = n % m;
  long long ex = m - 2;
  long long ans = 1;
  while (ex) {
    if (ex & 1) ans = ans * pw % m;
    pw = pw * pw % m;
    ex >>= 1;
  }
  return ans % 1000000007;
}
long long pw(long long a, long long n) {
  long long pw = a;
  long long ex = n;
  long long ans = 1;
  while (ex) {
    if (ex & 1) ans = ans * pw;
    pw = pw * pw;
    ex >>= 1;
  }
  return ans;
}
long long pwm(long long a, long long n) {
  long long pw = a % 1000000007;
  long long ex = n;
  long long ans = 1;
  while (ex) {
    if (ex & 1) ans = ans * pw % 1000000007;
    pw = pw * pw % 1000000007;
    ex >>= 1;
  }
  return ans % 1000000007;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long max(long long a, long long b) { return a > b ? a : b; }
long long min(long long a, long long b) { return a < b ? a : b; }
long long Abs(long long x) { return x > 0 ? x : (-x); }
long long Ceil(long long a, long long b) { return a / b + (a % b != 0); }
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
long long X[] = {2, 1, -1, -2, -2, -1, 1, 2};
long long Y[] = {1, 2, 2, 1, -1, -2, -2, -1};
void swap(long long *a, long long *b) {
  long long t = *a;
  *a = *b;
  *b = t;
}
struct comp {
  bool operator()(const pair<long long, long long> &a,
                  const pair<long long, long long> &b) const {
    if (a.first > b.first) return true;
    if (a.first < b.first) return false;
    if (a.second < b.second) return true;
    return false;
  }
};
long long n, x, y;
vector<long long> adj[100005];
vector<long long> prog;
long long dfs(long long child, long long parent) {
  long long sz = 1;
  for (long long subC : adj[child]) {
    if (subC != parent) {
      long long count = dfs(subC, child);
      prog.push_back(count * (n - count));
      sz += count;
    }
  }
  return sz;
}
void solve() {
  prog.clear();
  for (long long i = 0; i <= n; ++i) {
    adj[i].clear();
  }
  cin >> n;
  for (long long i = 0; i < n - 1; ++i) {
    cin >> x >> y;
    x--;
    y--;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  dfs(0, -1);
  sort(prog.begin(), prog.end(), greater<long long>());
  long long m;
  cin >> m;
  vector<long long> fm(m);
  for (long long i = 0; i < m; ++i) {
    cin >> fm[i];
  }
  for (long long i = m; i < n - 1; ++i) {
    fm.push_back(1);
  }
  sort(fm.begin(), fm.end(), greater<long long>());
  long long ans = 0;
  if (fm.size() == n - 1) {
    for (long long i = 0; i < n - 1; ++i) {
      ans = (ans % 1000000007 + (fm[i] * prog[i]) % 1000000007) % 1000000007;
    }
  } else {
    long long z = fm.size() - n + 1, x = 0;
    for (long long i = 0; i < fm.size(); ++i) {
      if (i < z)
        fm[i + 1] = (fm[i + 1] % 1000000007 * fm[i] % 1000000007) % 1000000007;
      else
        ans =
            (ans % 1000000007 + (fm[i] * prog[x++]) % 1000000007) % 1000000007;
    }
  }
  cout << ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long tst = 1;
  cin >> tst;
  while (tst--) {
    solve();
    cout << "\n";
  }
}
