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<vector<long long int>> v;
long long int n;
vector<long long int> value;
long long int size1[100005];
void count(long long int src, long long int parent) {
  size1[src] = 1;
  for (auto x : v[src]) {
    if (x != parent) {
      count(x, src);
      value.push_back(size1[x] * (n - size1[x]));
      size1[src] += size1[x];
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    v.clear();
    value.clear();
    memset(size1, (long long int)0, sizeof size1);
    cin >> n;
    v.assign(n + 1, vector<long long int>());
    for (long long int i = 0, a, b; i < n - 1; i++) {
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    long long int m1, mod = 1000000007, ans = 0;
    cin >> m1;
    vector<long long int> fact;
    for (long long int i = 0; i < m1; i++) {
      long long int temp;
      cin >> temp;
      fact.push_back(temp);
    }
    count(1, -1);
    while (fact.size() < n - 1) fact.push_back(1);
    m1 = fact.size();
    sort(fact.begin(), fact.end());
    sort(value.begin(), value.end());
    for (long long int i = 0; i < n - 2; i++) {
      ans = (ans + ((value[i] % mod) * (fact[i] % mod)) % mod) % mod;
    }
    long long int left = 1;
    for (long long int i = n - 2; i < m1; i++) {
      left = ((left % mod) * (fact[i] % mod)) % mod;
    }
    ans = (ans + (left * value[n - 2]) % mod) % mod;
    cout << ans << endl;
  }
  return 0;
}
