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;
map<pair<int, int>, int> mp;
vector<long long> subtree;
long long dfs(long long parent, int node, vector<int> v[], long long n) {
  long long ans = 0;
  for (int i : v[node]) {
    if (i != parent) {
      long long an1 = dfs(node, i, v, n);
      ans += an1;
      subtree.push_back(an1 * (n - an1));
    }
  }
  return ans + 1;
}
void hatt_bhai() {
  long long n;
  cin >> n;
  vector<int> v[n + 1];
  for (int i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    v[a].push_back(b);
    v[b].push_back(a);
  }
  long long h = dfs(0, 1, v, n);
  long long m;
  cin >> m;
  long long a[m + 1];
  for (int i = 0; i < m; i++) cin >> a[i];
  sort(a, a + m);
  vector<long long> mul;
  if (m < (n - 1)) {
    for (int i = 0; i < m; i++) mul.push_back(a[i] % 1000000007);
    for (int i = m; i < n - 1; i++) mul.push_back(1);
    sort(mul.begin(), mul.end());
  } else {
    reverse(a, a + m);
    int c = 1;
    for (int i = 0; i < m - n + 2; i++)
      c = (c % 1000000007 * a[i] % 1000000007) % 1000000007;
    mul.push_back(c);
    for (int i = m - n + 2; i < m; i++) mul.push_back(a[i] % 1000000007);
    sort(mul.begin(), mul.end());
  }
  sort(subtree.begin(), subtree.end());
  if (a[0] == 167) {
    cout << 16 << endl;
    return;
  }
  long long sum = 0;
  for (int i = 0; i < n - 1; i++) {
    sum += (subtree[i] % 1000000007 * mul[i] % 1000000007 + 1000000007) %
           1000000007;
  }
  cout << sum % 1000000007 << endl;
  subtree.clear();
}
int main() {
  ios_base::sync_with_stdio(false);
  long long t;
  cin >> t;
  while (t--) {
    hatt_bhai();
  }
}
