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;
vector<long long> graph[100000 + 5];
vector<long long> cont;
long long dfs(long long u, long long p, long long n) {
  long long cnt = 0;
  for (auto z : graph[u]) {
    if (z != p) {
      long long temp = dfs(z, u, n);
      cont.push_back(1LL * temp * (n - temp));
      cnt += temp;
    }
  }
  cnt += 1;
  return cnt;
}
inline long long mul(long long a, long long b, long long mod = 1000000007) {
  long long c = 1LL * a * b;
  if (c >= mod) {
    c %= mod;
  }
  return c;
}
inline long long add(long long a, long long b, long long mod = 1000000007) {
  long long c = a + b;
  if (c >= mod) {
    c %= mod;
  }
  return c;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t-- > 00) {
    long long n, u, v, i, m;
    cin >> n;
    for (i = 1; i <= n - 1; ++i) {
      cin >> u >> v;
      graph[u].push_back(v);
      graph[v].push_back(u);
    }
    dfs(1, 0, n);
    cin >> m;
    long long p[m + 5];
    for (i = 0; i < m; ++i) {
      cin >> p[i];
    }
    sort(p, p + m);
    sort(cont.begin(), cont.end());
    assert(cont.size() == n - 1);
    long long ans = 0;
    if (n - 1 >= m) {
      for (i = n - 2; m > 0 and i >= 0; --i, --m) {
        ans = add(ans, mul(cont[i], p[m - 1]));
      }
      for (; i >= 0; --i) {
        ans = add(ans, cont[i]);
      }
    } else {
      for (i = 0; i < n - 2; ++i) {
        ans = add(ans, mul(cont[i], p[i]));
      }
      long long temp = cont[i];
      for (; i < m; ++i) {
        temp = mul(temp, p[i]);
      }
      ans = add(ans, temp);
    }
    cout << ans << "\n";
    for (i = 1; i <= n; ++i) {
      graph[i].clear();
    }
    cont.clear();
  }
  return 0;
}
