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 inf = 1000000000000000000;
const long long mod = 1000000007;
long long n, x, y, m, a[60005], perv, uc, ans, tec, t;
vector<long long> edges[100005], myvec;
bool used[100005];
long long dfs(long long v) {
  long long kol = 1;
  used[v] = true;
  for (long long i = 0; i < (long long)edges[v].size(); ++i)
    if (!used[edges[v][i]]) {
      kol += dfs(edges[v][i]);
    }
  myvec.push_back(kol * (n - kol));
  return kol;
}
bool fcmp(long long a, long long b) { return (a > b); }
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (long long o = 1; o <= t; ++o) {
    cin >> n;
    for (long long i = 1; i <= n; ++i) edges[i].clear();
    for (long long i = 1; i < n; ++i) {
      cin >> x >> y;
      edges[x].push_back(y);
      edges[y].push_back(x);
    }
    for (long long i = 1; i <= n; ++i) used[i] = false;
    used[1] = true;
    myvec.clear();
    for (long long i = 0; i < (long long)edges[1].size(); ++i)
      x = dfs(edges[1][i]);
    sort(myvec.begin(), myvec.end(), fcmp);
    cin >> m;
    for (long long i = 1; i <= m; ++i) cin >> a[i];
    sort(a + 1, a + m + 1, fcmp);
    if (m > n - 2)
      perv = m - (n - 2);
    else
      perv = 1;
    ans = 0;
    uc = 1;
    for (long long i = 0; i < (long long)myvec.size(); ++i) {
      tec = 1;
      if (uc > m) {
        ans += myvec[i];
        ans %= mod;
        continue;
      }
      if (uc == 1) {
        for (long long j = 1; j <= perv; ++j) {
          tec *= a[uc];
          tec %= mod;
          ++uc;
        }
      } else {
        tec *= a[uc];
        tec %= mod;
        ++uc;
      }
      tec *= myvec[i];
      tec %= mod;
      ans += tec;
      ans %= mod;
    }
    cout << ans << '\n';
  }
  return 0;
}
