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 int mod = 1e9 + 7;
const long long int cmod = 998244353;
const long long int N = 3e5 + 5;
const long long int inf = 1e16 + 2;
vector<long long int> graph[N];
long long int p[N], s[N];
long long int dfs(long long int idx, long long int par) {
  for (auto x : graph[idx]) {
    if (x == par) continue;
    p[idx] += dfs(x, idx);
  }
  p[idx]++;
  return p[idx];
}
long long int n;
void dfs2(long long int idx, long long int par) {
  s[idx] = (n - p[idx]) * (p[idx]);
  for (auto x : graph[idx]) {
    if (x == par) continue;
    dfs2(x, idx);
  }
}
void solve(long long int Case) {
  cin >> n;
  for (long long int i = 0; i < n; i++) p[i] = 0, s[i] = 0, graph[i].clear();
  for (long long int i = 0; i < n - 1; i++) {
    long long int a, b;
    cin >> a >> b;
    a--, b--;
    graph[a].push_back(b);
    graph[b].push_back(a);
  }
  long long int k;
  cin >> k;
  priority_queue<long long int> K;
  for (long long int i = 0; i < k; i++) {
    long long int a;
    cin >> a;
    K.push(a);
  }
  if (k < n - 1) {
    for (long long int i = k; i < n - 1; i++) K.push(1);
  } else {
    long long int a = 1;
    while (K.size() >= n - 1) {
      a = (a * K.top()) % mod;
      K.pop();
    }
    K.push(a);
  }
  dfs(0, -1);
  dfs2(0, -1);
  vector<long long int> v;
  for (long long int i = 1; i < n; i++) v.push_back(s[i]);
  sort(v.rbegin(), v.rend());
  long long int ans = 0;
  for (auto x : v) {
    long long int a = K.top();
    ans = (ans + ((x) % mod * a % mod) % mod) % mod;
    K.pop();
  }
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int testcase = 1;
  cin >> testcase;
  long long int Case = 1;
  while (testcase--) {
    solve(Case++);
  }
  return 0;
}
