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;
vector<long long> vec[100005];
long long num[60005];
vector<long long> cnt;
long long size[100005];
long long n;
void dfs(long long u, long long fa) {
  size[u] = 1;
  for (long long i = 0; i < vec[u].size(); i++) {
    if (vec[u][i] == fa) continue;
    dfs(vec[u][i], u);
    size[u] += size[vec[u][i]];
  }
  cnt.push_back((n - size[u]) * size[u]);
}
bool cmp(long long a, long long b) { return a > b; }
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 0; i <= n; i++) vec[i].clear();
    cnt.clear();
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      scanf("%lld%lld", &u, &v);
      vec[u].push_back(v);
      vec[v].push_back(u);
    }
    dfs(1, 0);
    sort(cnt.begin(), cnt.end(), cmp);
    long long m;
    scanf("%lld", &m);
    for (long long i = 0; i < m; i++) scanf("%lld", &num[i]);
    sort(num, num + m, cmp);
    long long p = 0;
    n--;
    long long sum = 0;
    if (m - n > 0) {
      while (p < m - n) {
        num[p + 1] = num[p] * num[p + 1] % mod;
        p++;
      }
      for (int i = p; i < m; i++) {
        sum = (sum + num[i] * cnt[i - p] % mod) % mod;
      }
    } else {
      for (long long i = 0; i < n; i++) {
        if (i >= m)
          sum = (sum + cnt[i]) % mod;
        else
          sum = (sum + num[i] * cnt[i] % mod) % mod;
      }
    }
    printf("%lld\n", sum);
  }
  return 0;
}
