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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
template <typename T, typename U>
static inline void amin(T& x, U y) {
  if (y < x) x = y;
}
template <typename T, typename U>
static inline void amax(T& x, U y) {
  if (x < y) x = y;
}
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; }
const long long N = 1e5 + 10;
vector<long long> adj[N];
long long n;
vector<long long> vals;
long long dfs(long long v, long long p, long long up = 0) {
  long long z = 0;
  long long sm = 1;
  for (auto i : adj[v]) {
    if (i == p) continue;
    long long t = dfs(i, v, up + 1);
    sm += t;
    long long left = n - t;
    z = t * left;
    vals.push_back(z);
  }
  return sm;
}
long long solve() {
  cin >> n;
  vals.clear();
  for (long long i = 1; i < n + 1; i++) adj[i].clear();
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  long long m;
  cin >> m;
  long long p[m];
  for (long long i = 0; i < m; ++i) cin >> p[i];
  sort(p, p + m);
  vector<long long> primes;
  if (m > n - 1) {
    long long x = 1;
    for (long long j = n - 2; j < m; j++) {
      x = (x * p[j]) % 1000000007;
    }
    primes.push_back(x);
    for (long long j = n - 3; j >= 0; j--) {
      primes.push_back(p[j]);
    }
  } else {
    for (long long j = m - 1; j >= 0; j--) primes.push_back(p[j]);
    while (primes.size() < n) primes.push_back(1);
  }
  dfs(1, -1);
  sort(vals.begin(), vals.end(), greater<long long>());
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) {
    ans += ((vals[i] % 1000000007) * primes[i]) % 1000000007;
    ans %= 1000000007;
  }
  cout << ans << '\n';
  return 0;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
