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 int N = 200010;
long long t, n, m, p[N], d[N], a[N];
vector<long long> adj[N], v;
long long dfs(long long s, long long pre) {
  p[s] = 1;
  for (auto u : adj[s]) {
    if (u != pre) {
      p[s] += dfs(u, s);
    }
  }
  if (s > 1) v.push_back(p[s] * (n - p[s]));
  return p[s];
}
void rest() {
  for (int i = 1; i <= n; i++) {
    adj[i].clear();
  }
  v.clear();
}
int solve() {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (int i = 0; i < n - 1; i++) {
      long long u, v;
      scanf("%lld%lld", &u, &v);
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    scanf("%lld", &m);
    for (int i = 0; i < m; i++) {
      scanf("%lld", &a[i]);
    }
    sort(a, a + m);
    long long ret = dfs(1, 1), res = 0;
    sort(v.begin(), v.end());
    long long sz = v.size();
    if (m <= sz) {
      int i, j;
      for (i = m - 1, j = sz - 1; i >= 0; i--, j--) {
        res = (res + (v[j] * a[i]) % 1000000007) % 1000000007;
      }
      while (j >= 0) {
        res = (res + v[j]) % 1000000007;
        j--;
      }
    } else {
      int i, j;
      for (i = 0, j = 0; j < sz - 1; i++, j++) {
        res = (res + (v[j] * a[i]) % 1000000007) % 1000000007;
      }
      long long x = 1;
      while (i < m) {
        x = (x * a[i]) % 1000000007;
        i++;
      }
      res = (res + (x * v[sz - 1]) % 1000000007) % 1000000007;
    }
    printf("%lld\n", res);
    rest();
  }
  return 0;
}
int main() { return solve(); }
