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 n, m, a[100010], ans[100010], tot;
vector<long long> son[100010];
long long out, mod = 1e9 + 7;
long long cnt[100010], u, v;
void dfs(long long x, long long fa) {
  for (vector<long long>::iterator it = son[x].begin(); it != son[x].end();
       it++) {
    long long to = *it;
    if (to == fa) continue;
    dfs(to, x);
    ans[tot++] = cnt[to] * (n - cnt[to]);
    cnt[x] += cnt[to];
  }
}
inline void solve(int T) {
  cin >> n;
  tot = 1;
  for (register int i = 1; i <= n; i++) {
    son[i].clear();
    cnt[i] = 1;
  }
  for (register int i = 1; i <= n - 1; i++) {
    cin >> u >> v;
    son[u].push_back(v);
    son[v].push_back(u);
  }
  cin >> m;
  for (register int i = 1; i <= m; i++) cin >> a[i];
  if (m < n) {
    for (register int i = m + 1; i <= n - 1; i++) a[i] = 1;
    sort(a + 1, a + n);
  } else {
    sort(a + 1, a + m + 1);
    for (register int i = n; i <= m; i++) a[n - 1] = a[n - 1] * a[i] % mod;
  }
  dfs(1, 0);
  sort(ans + 1, ans + n);
  out = 0;
  for (register int i = 1; i <= n - 1; i++)
    out = (out + ans[i] % mod * a[i] % mod) % mod;
  cout << out << endl;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int T = 1;
  cin >> T;
  for (register int i = 1; i <= T; i++) solve(i);
}
