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 MOD = (int)1e9 + 7;
int t, n, m, x, y;
vector<int> g[100005];
int v[100005], w[100005], dp[100005];
void dfs(int nod, int tata = 0) {
  dp[nod] = 1;
  for (auto &fiu : g[nod]) {
    if (fiu != tata) dfs(fiu, nod), dp[nod] += dp[fiu];
  }
}
bool comp(int a, int b) {
  return 1LL * dp[a] * (n - dp[a]) > 1LL * dp[b] * (n - dp[b]);
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  for (; t; t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) g[i].clear(), w[i] = i;
    for (int i = 1; i < n; i++) {
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    dfs(1);
    cin >> m;
    for (int i = 1; i <= m; i++) cin >> v[i];
    sort(v + 1, v + m + 1);
    sort(w + 1, w + n + 1, comp);
    int j = m, ans = 0, p = 1;
    while (j > n - 1) p = 1LL * p * v[j] % MOD, j--;
    p = 1LL * p * v[j] % MOD;
    ans = (ans + 1LL * dp[w[1]] * (n - dp[w[1]]) * p) % MOD, j--;
    for (int i = 2; i <= n; i++) {
      if (!j)
        ans = (ans + 1LL * dp[w[i]] * (n - dp[w[i]])) % MOD;
      else
        ans = (ans + 1LL * dp[w[i]] * (n - dp[w[i]]) * v[j]) % MOD, j--;
    }
    cout << ans << "\n";
  }
  return 0;
}
