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 = 1e9 + 7;
long long p[60005];
long long gt[100005];
vector<int> E[100005];
long long sz[100005];
int tot;
long long ad(long long a, long long b) {
  return (((a % mod + b % mod) % mod) + mod) % mod;
}
long long ap(long long a, long long b) { return ((a % mod) * (b % mod)) % mod; }
int n;
void dfs(int u, int fa) {
  sz[u] = 1;
  for (int i = 0; i <= E[u].size() - 1; i++) {
    int v = E[u][i];
    if (v == fa) continue;
    dfs(v, u);
    sz[u] += sz[v];
  }
  for (int i = 0; i <= E[u].size() - 1; i++) {
    int v = E[u][i];
    if (v == fa) continue;
    gt[++tot] = 1ll * (n - sz[v]) * sz[v];
  }
}
int main(int argc, char** argv) {
  int T;
  T = 1;
  scanf("%d", &T);
  while (T--) {
    tot = 0;
    int m;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) E[i].clear();
    for (int i = 1; i <= n - 1; i++) {
      int u, v;
      scanf("%d", &u);
      scanf("%d", &v);
      E[u].push_back(v);
      E[v].push_back(u);
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) scanf("%lld", &p[i]);
    sort(p + 1, p + 1 + m);
    dfs(1, -1);
    sort(gt + 1, gt + 1 + tot);
    int j = 0;
    long long ans = 0;
    while (m > tot) {
      p[m - 1] = ap(p[m], p[m - 1]);
      m--;
    }
    for (int i = 1; i <= tot; i++) {
      if (tot - i + 1 <= m) {
        ans = ad(ans, ap(p[++j], gt[i]));
      } else
        ans = ad(ans, gt[i]);
    }
    cout << ans << endl;
  }
  return 0;
}
