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 mo = 1000000007;
const int N = 100005;
int n, sz[N], m;
int q[N * 50], qq[N * 50];
long long b[N * 50];
vector<int> e[N];
void dfs(int x, int fa) {
  sz[x] = 1;
  for (auto i : e[x])
    if (i != fa) {
      dfs(i, x);
      sz[x] += sz[i];
      b[++*b] = 1ll * sz[i] * (n - sz[i]);
    }
}
void solve() {
  *b = 0;
  scanf("%d", &n);
  for (int i = (int)(1); i <= (int)(n); i++) e[i].resize(0);
  for (int i = (int)(1); i <= (int)(n - 1); i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  *q = 0;
  scanf("%d", &m);
  for (int i = (int)(1); i <= (int)(m); i++) {
    int x;
    scanf("%d", &x);
    for (int j = 2; j * j <= x; j++)
      for (; x % j == 0; x /= j, q[++*q] = j)
        ;
    if (x != 1) q[++*q] = x;
  }
  sort(q + 1, q + *q + 1);
  if (*q < n) {
    for (int i = (int)(1); i <= (int)(n - 1); i++) qq[i] = 1;
    for (int i = (int)(1); i <= (int)(*q); i++) qq[n - 1 - (*q - i)] = q[i];
  } else {
    for (int i = (int)(1); i <= (int)(n - 1); i++) qq[i] = 1;
    for (int i = (int)(1); i <= (int)(n - 1); i++) qq[i] = q[i];
    for (int i = (int)(n); i <= (int)(*q); i++)
      qq[n - 1] = 1ll * qq[n - 1] * q[i] % mo;
  }
  dfs(1, 0);
  sort(b + 1, b + n);
  int ans = 0;
  for (int i = (int)(1); i <= (int)(n - 1); i++)
    ans = (ans + b[i] % mo * qq[i]) % mo;
  printf("%d\n", ans);
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) solve();
}
