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 double pi = 3.1415926535897932384626433;
const int mod = 1000000007;
const long long inf = 3e18;
vector<vector<long long> > adj(100001);
bool vis[100001];
long long st[100001];
vector<long long> v;
void dfs(long long x, long long n) {
  vis[x] = true;
  st[x] = 1;
  for (auto it : adj[x]) {
    if (!vis[it]) {
      dfs(it, n);
      st[x] += st[it];
    }
  }
  v.push_back(st[x] * (n - st[x]));
}
void solve() {
  long long n;
  cin >> n;
  long long ans = 0;
  memset(st, 0, sizeof(st));
  memset(vis, 0, sizeof(vis));
  v.clear();
  for (long long i = 1; i <= 100000; i++) {
    adj[i].clear();
  }
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, n);
  sort(v.rbegin(), v.rend());
  long long m;
  cin >> m;
  vector<long long> pf(m);
  for (auto &it : pf) {
    cin >> it;
  }
  while (m < n - 1) {
    pf.push_back(1LL);
    m++;
  }
  sort(pf.rbegin(), pf.rend());
  if (m > n - 1) {
    long long i, x;
    for (i = 1; i <= m - n + 1; i++) {
      pf[0] = (pf[0] % mod * pf[i] % mod);
      pf[0] = pf[0] % mod;
    }
    for (x = 1; i < m; i++, x++) {
      pf[x] = pf[i];
    }
  }
  sort(v.rbegin(), v.rend());
  for (long long i = 0; i < n - 1; i++) {
    ans = ans % mod + (pf[i] % mod * v[i] % mod) % mod;
    ans = ans % mod;
  }
  cout << ans << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}
