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;
long long cnt, head[100005], sz[100005], ans, n, m, mx, T, x, y, tp, a[100005];
priority_queue<long long, vector<long long>, less<long long> > q;
struct {
  long long to, nx;
} e[100005 << 1];
void ad(long long x, long long y) {
  cnt++;
  e[cnt].to = y;
  e[cnt].nx = head[x];
  head[x] = cnt;
}
void dfs(long long k, long long fa) {
  sz[k] = 1;
  for (long long i = head[k]; i; i = e[i].nx)
    if (e[i].to != fa) dfs(e[i].to, k), sz[k] += sz[e[i].to];
}
void dfs1(long long k, long long fa) {
  for (long long i = head[k]; i; i = e[i].nx)
    if (e[i].to != fa) {
      q.push(sz[e[i].to] * (n - sz[e[i].to]));
      dfs1(e[i].to, k);
    }
}
signed main() {
  cin >> T;
  while (T--) {
    cin >> n;
    cnt = ans = mx = 0;
    for (long long i = 1; i <= n; i++) head[i] = 0;
    for (long long i = 1; i < n; i++) {
      cin >> x >> y;
      ad(x, y);
      ad(y, x);
    }
    dfs(1, 0);
    dfs1(1, 0);
    cin >> m;
    for (long long i = 1; i <= m; i++) {
      cin >> a[i];
    }
    sort(a + 1, a + m + 1);
    n--;
    if (m > n) {
      for (long long i = n + 1; i <= m; i++) (a[n] *= a[i]) %= 1000000007;
      m = n;
    }
    while (!q.empty()) {
      tp = q.top();
      q.pop();
      (ans += max(1ll, a[m--]) * tp % 1000000007) %= 1000000007;
      if (m < 0) m = 0;
    }
    cout << ans << endl;
  }
}
