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;
const long long int M = 1000000007;
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res % p;
}
long long int modinver(long long int a, long long int b) {
  return (power(a, 1, M) * power(b, M - 2, M)) % M;
}
void SieveOfEratosthenes(int n) {
  bool prime[10005];
  memset(prime, true, sizeof(prime));
  for (int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (int i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
std::vector<long long int> v[100002];
std::vector<long long int> edge;
long long int vis[100002];
long long int n;
long long int dfs(int x) {
  vis[x] = 1;
  long long int co = 1;
  for (auto i : v[x]) {
    if (vis[i] == 0) {
      long long int aa = dfs(i);
      edge.push_back(aa * (n - aa));
      co += aa;
    }
  }
  return co;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i <= n; ++i) {
      vis[i] = 0;
      edge.clear();
      v[i].clear();
    }
    for (int i = 0; i < n - 1; ++i) {
      long long int a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    long long int m;
    cin >> m;
    std::vector<long long int> pr(m);
    for (int i = 0; i < m; ++i) {
      cin >> pr[i];
    }
    sort(pr.begin(), pr.end());
    dfs(1);
    if (m < n - 1) {
      for (int i = 0; i < n - 1 - m; ++i) {
        pr.push_back(1);
      }
      sort(pr.begin(), pr.end());
    } else if (m > n - 1) {
      long long int aa = pr[m - 1];
      for (int i = m - 2; i >= n - 2; --i) {
        aa = (pr[i] * aa) % M;
      }
      pr[n - 2] = aa;
    }
    long long int ans = 0;
    sort(edge.begin(), edge.end());
    for (int i = 0; i < n - 1; ++i) {
      ans = ans + ((edge[i] % M) * pr[i]) % M;
      ans = ans % M;
    }
    cout << ans << '\n';
  }
}
