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 MX = 1e5 + 5;
const long long int MOD = 1e9 + 7;
vector<int> adj[MX], child[MX];
int t, n, u, v, m;
bool visited[MX];
long long int p[MX], w[MX], SIZE[MX];
void getSize(int pos) {
  SIZE[pos] = 1;
  for (int next : child[pos]) {
    getSize(next);
    SIZE[pos] += SIZE[next];
  }
  w[pos - 1] = SIZE[pos] * (n - SIZE[pos]);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) child[i].clear(), adj[i].clear();
    for (int i = 0; i < n - 1; i++) {
      scanf("%d %d", &u, &v);
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) scanf("%lld", &p[i]);
    for (int i = m + 1; i < n; i++) p[i] = 1;
    m = max(m, n - 1);
    sort(p + 1, p + m + 1);
    for (int i = m; i >= n; i--) p[n - 1] = (p[i] * p[n - 1]) % MOD;
    fill(visited + 1, visited + n + 1, false);
    queue<int> q;
    q.push(1), visited[1] = true;
    while (!q.empty()) {
      int pos = q.front();
      q.pop();
      for (int next : adj[pos]) {
        if (visited[next]) continue;
        visited[next] = true;
        child[pos].push_back(next);
        q.push(next);
      }
    }
    getSize(1);
    sort(w + 1, w + n);
    long long int ans = 0;
    for (int i = 1; i < n; i++) ans = (ans + (w[i] % MOD) * p[i]) % MOD;
    printf("%lld\n", ans);
  }
}
