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 mod = 1000000007;
vector<long long> V[1000005];
long long vis[1000005];
long long X[1000005];
long long Y[1000005];
void dfs(int u, int fa) {
  vis[u] = 1;
  for (int i = 0; i < V[u].size(); i++) {
    int v = V[u][i];
    if (v != fa) {
      dfs(v, u);
      vis[u] += vis[v];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long T;
  cin >> T;
  while (T--) {
    long long n;
    cin >> n;
    for (int i = 1; i <= n; i++) {
      V[i].clear();
      vis[i] = 0;
    }
    for (int i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      X[i] = x;
      Y[i] = y;
      V[x].push_back(y);
      V[y].push_back(x);
    }
    dfs(1, 1);
    priority_queue<long long> Q;
    for (int i = 1; i < n; i++) {
      long long x = 0;
      long long y = 0;
      long long cnt = 0;
      x = vis[X[i]];
      y = vis[Y[i]];
      if (x < y) {
        cnt = x * (n - x);
      } else {
        cnt = y * (n - y);
      }
      Q.push(cnt);
    }
    long long m;
    cin >> m;
    vector<long long> W;
    for (int i = 1; i <= m; i++) {
      long long x;
      cin >> x;
      W.push_back(x);
    }
    sort(W.begin(), W.end());
    while (W.size() < (n - 1)) {
      W.push_back(1);
    }
    while (W.size() > (n - 1)) {
      long long e = W.back();
      W.pop_back();
      e = e * W.back();
      W.pop_back();
      e = e % mod;
      W.push_back(e);
    }
    sort(W.begin(), W.end());
    long long sum = 0;
    while (Q.size() > 0) {
      long long x = Q.top();
      Q.pop();
      x = x * W.back();
      W.pop_back();
      x = x % mod;
      sum += x;
      sum = sum % mod;
    }
    cout << sum << "\n";
  }
  return 0;
}
