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>
const long long N = 2e5 + 7;
const long long M = 2e4 + 5;
const double eps = 1e-8;
const long long mod = 1e9 + 7;
const long long inf = 0x7fffffff;
const double pi = 3.1415926;
using namespace std;
vector<long long> v[N], num;
long long n;
long long dfs(long long x, long long fa) {
  long long sum = 1;
  for (long long i = 0; i < v[x].size(); i++) {
    if (v[x][i] == fa) continue;
    sum += dfs(v[x][i], x);
  }
  if (x != 1) num.push_back((n - sum) * sum);
  return sum;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (long long i = 0; i <= n; i++) v[i].clear();
    num.clear();
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y);
      v[y].push_back(x);
    }
    long long m;
    cin >> m;
    vector<long long> p;
    for (long long i = 0; i < m; i++) {
      long long c;
      cin >> c;
      p.push_back(c);
    }
    dfs(1, -1);
    if (m < n - 1) {
      for (long long i = 0; i < n - 1 - m; i++) p.push_back(1);
    }
    sort(num.begin(), num.end());
    sort(p.begin(), p.end());
    long long ans = 0;
    if (m <= n - 1) {
      for (long long i = 0; i < n - 1; i++) {
        ans += (num[i] * p[i]) % mod;
        ans %= mod;
      }
    } else {
      for (long long i = 0; i < n - 2; i++) {
        ans += (num[i] * p[i]) % mod;
        ans %= mod;
      }
      long long add = 1;
      for (long long i = n - 2; i < m; i++) {
        add = (add * p[i]) % mod;
      }
      ans += (add * num[n - 2]) % mod;
      ans %= mod;
    }
    cout << ans << endl;
  }
  return 0;
}
