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 base = (int)(1e9 + 7);
const int N = 1e5 + 500;
int n;
long long sz[N];
vector<long long> vec;
vector<int> g[N];
long long mul(long long x, long long y) {
  x *= y;
  return x % base;
}
void dfs(int v, int pr) {
  sz[v] = 1;
  for (int u : g[v]) {
    if (u == pr) continue;
    dfs(u, v);
    sz[v] += sz[u];
    vec.push_back(sz[u] * (n - sz[u]));
  }
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int v, u;
    cin >> v >> u;
    --v, --u;
    g[v].push_back(u);
    g[u].push_back(v);
  }
  dfs(0, -1);
  sort(vec.rbegin(), vec.rend());
  int m;
  cin >> m;
  vector<int> a(m);
  for (int &i : a) cin >> i;
  sort(a.begin(), a.end());
  --m;
  while (m > n - 2) a[m - 1] = mul(a[m - 1], a[m]), --m;
  long long ans = 0;
  for (int i = 0; i < n - 1; ++i)
    if (m < 0)
      ans = (ans + vec[i]) % base;
    else
      ans = (ans + mul(vec[i], a[m])) % base, --m;
  cout << ans << "\n";
  for (int i = 0; i < n; ++i) g[i].clear();
  vec.clear();
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
