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 maxn = 2e5 + 10;
const int maxm = 5e5 + 10;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
int dir[4][2] = {{-1, 0}, {0, 1}, {1, 0}, {0, -1}};
long long a[maxn];
int t, n, m;
vector<int> G[maxn];
vector<long long> v;
long long sz[maxn], ans;
void init() {
  ans = 0;
  v.clear();
  for (int i = 1; i <= n; i++) G[i].clear();
  memset(sz, 0, sizeof(sz));
}
void dfs(int u, int fa) {
  sz[u] = 1ll;
  for (auto i : G[u]) {
    if (i == fa) continue;
    dfs(i, u);
    sz[u] += sz[i];
    v.push_back((n - sz[i]) * sz[i]);
  }
}
void solve() {
  init();
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d", &u), scanf("%d", &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%lld", &a[i]);
  sort(a + 1, a + m + 1);
  while (m > n - 1) {
    a[m - 1] = (a[m - 1] * a[m]) % mod;
    m--;
  }
  dfs(1, -1);
  sort(v.begin(), v.end());
  int k = v.size();
  for (int i = k - 1; i >= 0; i--) {
    if (m == 0)
      ans = (ans + v[i]) % mod;
    else
      ans = (ans + v[i] % mod * a[m] % mod) % mod, m--;
  }
  printf("%lld\n", ans);
}
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    solve();
  }
  return 0;
}
