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 = 1e5 + 10;
vector<long long> g[maxn];
vector<long long> vp;
long long n, m;
long long dfs(long long u, long long p) {
  long long ans = 1;
  for (long long v : g[u]) {
    if (v == p) {
      continue;
    }
    long long x = dfs(v, u);
    vp.push_back(x * (n - x));
    ans += x;
  }
  return ans;
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  cin >> m;
  vector<long long> p(m);
  for (long long i = 0; i < m; ++i) {
    cin >> p[i];
  }
  sort(p.begin(), p.end());
  long long temp = dfs(1, -1);
  sort(vp.begin(), vp.end());
  long long res = 0;
  long long i = n - 2;
  if (m <= n - 1) {
    for (; i >= 0; --i) {
      if (p.size() == 0) {
        break;
      }
      res += (vp[i] * p.back()) % 1000000007;
      res %= 1000000007;
      p.pop_back();
    }
    while (i >= 0) {
      res += vp[i];
      res %= 1000000007;
      --i;
    }
  } else {
    long long temp = 1;
    long long z = (m - (n - 1));
    z++;
    while (z--) {
      temp *= p.back();
      temp %= 1000000007;
      p.pop_back();
    }
    res += (vp[n - 2] * temp) % 1000000007;
    res %= 1000000007;
    i = n - 3;
    for (; i >= 0; --i) {
      res += (vp[i] * p.back()) % 1000000007;
      res %= 1000000007;
      p.pop_back();
    }
  }
  cout << res % 1000000007 << "\n";
  for (long long i = 1; i <= n; ++i) {
    g[i].clear();
  }
  vp.clear();
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  long long t;
  cin >> t;
  for (long long i = 1; i <= t; ++i) {
    solve();
  }
}
