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 N = 1e5 + 9, mod = 1e9 + 7;
long long n, m, edge, weight[N];
vector<long long> k;
vector<long long> g[N];
bool vi[N];
long long dfs(long long x, long long pai) {
  vi[x] = true;
  long long w = 1;
  for (long long i = 0; i < (long long)g[x].size(); i++) {
    long long y = g[x][i];
    if (y != pai && !vi[y]) {
      w += dfs(y, x);
    }
  }
  weight[edge] = w * (n - w);
  edge++;
  return w;
}
int main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 0; i < n - 1; i++) {
      long long a, b;
      scanf("%lld %lld", &a, &b);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    dfs(1, -1);
    sort(weight, weight + n - 1);
    reverse(weight, weight + n - 1);
    for (int i = 0; i < edge; i++) weight[i] %= mod;
    scanf("%lld", &m);
    for (long long i = 0; i < m; i++) {
      long long x;
      scanf("%lld", &x);
      k.push_back(x);
    }
    sort(k.begin(), k.end());
    reverse(k.begin(), k.end());
    if (m <= n - 1) {
      long long ans = 0;
      while ((int)k.size() < n - 1) k.push_back(1);
      for (long long i = 0; i < n; i++) {
        ans = (ans + weight[i] * k[i]) % mod;
      }
      printf("%lld\n", ans);
    } else {
      n--;
      long long exc = m - n, ans = 0;
      long long prod = 1;
      for (long long i = 0; i < exc + 1; i++) prod = (prod * k[i]) % mod;
      ans = (ans + prod * weight[0]);
      for (long long i = 1; i < n; i++) {
        ans = (ans + k[i + exc] * weight[i]) % mod;
      }
      printf("%lld\n", ans);
    }
    for (int i = 1; i <= n + 1; i++) g[i].clear();
    memset(vi, 0, sizeof vi);
    memset(weight, 0, sizeof weight);
    k.clear();
    edge = 0;
  }
  return 0;
}
