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 maxn = 2e5 + 50, N = 60000 + 20, mod = 1e9 + 7, INF = 1e9 + 1,
                lg = 32;
long long sz[maxn];
vector<long long> num, p;
vector<pair<long long, long long> > edges[maxn];
void dfs(long long v, long long par, long long n) {
  sz[v] = 1;
  for (auto i : edges[v]) {
    long long u = i.first, id = i.second;
    if (u != par) {
      dfs(u, v, n);
      sz[v] += sz[u];
      num.push_back(1ll * sz[u] * (n - sz[u]));
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n, m;
    cin >> n;
    num.clear();
    p.clear();
    for (long long i = 0; i < n; ++i) {
      edges[i].clear();
      sz[i] = 0;
    }
    for (long long i = 0; i < n - 1; ++i) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      edges[x].push_back({y, i});
      edges[y].push_back({x, i});
    }
    dfs(0, -1, n);
    cin >> m;
    for (long long i = 0; i < m; ++i) {
      long long x;
      cin >> x;
      p.push_back(x);
    }
    sort(num.begin(), num.end());
    if (m <= n - 1) {
      long long ans = 0;
      long long need = (n - 1) - m;
      for (long long i = 0; i < need; ++i) {
        p.push_back(1ll);
      }
      sort(p.begin(), p.end());
      for (long long i = 0; i < n - 1; ++i) {
        ans = (ans + ((num[i] % mod) * p[i])) % mod;
      }
      cout << ans << '\n';
    } else {
      long long ans = 0;
      sort(p.begin(), p.end());
      long long bad = m - (n - 1);
      bad++;
      vector<long long> w;
      w.clear();
      long long curr = 1;
      for (long long i = 0; i < bad; ++i) {
        curr = (curr * p[m - 1 - i]) % mod;
      }
      for (long long i = 0; i < m - bad; ++i) {
        w.push_back(p[i]);
      }
      sort(w.begin(), w.end());
      w.push_back(curr);
      for (long long i = 0; i < n - 1; ++i) {
        ans = (ans + ((num[i] % mod) * (w[i] % mod))) % mod;
      }
      cout << ans << '\n';
    }
  }
}
