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 tc[(long long)1e5 + 4], n;
vector<long long> wt;
void dfs(long long st, vector<long long> g[], long long par) {
  for (auto x : g[st])
    if (x != par) dfs(x, g, st);
  tc[st] = 1;
  for (auto x : g[st]) tc[st] += tc[x];
}
void dfs2(long long st, vector<long long> g[], long long par) {
  for (auto x : g[st])
    if (x != par) dfs2(x, g, st);
  for (auto x : g[st])
    if (x != par) wt.push_back(tc[x] * (n - tc[x]));
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  long long tt;
  cin >> tt;
  wt.clear();
  while (tt--) {
    wt.clear();
    long long m, a, b;
    cin >> n;
    vector<long long> g[n + 1];
    for (long long i = 1; i <= n; i++) tc[i] = 0;
    for (long long i = 1; i <= n - 1; i++) {
      cin >> a >> b;
      g[a].push_back(b), g[b].push_back(a);
    }
    cin >> m;
    vector<long long> pp(m);
    for (long long i = 0; i <= m - 1; i++) cin >> pp[i];
    dfs(1, g, -1);
    dfs2(1, g, -1);
    long long ans = 0;
    sort(wt.begin(), wt.end());
    reverse(wt.begin(), wt.end());
    sort(pp.begin(), pp.end());
    reverse(pp.begin(), pp.end());
    long long mi = 0;
    if (m >= n - 1) {
      long long prod = 1;
      while (m - mi >= n - 1) prod = (prod * pp[mi++]) % ((long long)(1e9 + 7));
      long long kpo = wt.size();
      ans = (ans + prod * wt[0]) % ((long long)(1e9 + 7));
      for (long long i = 1; i <= kpo - 1; i++)
        ans = (ans + pp[mi++] * wt[i]) % ((long long)(1e9 + 7));
      cout << ans << "\n";
    } else {
      while (m != wt.size()) pp.push_back(1), m++;
      long long kpo = wt.size();
      for (long long i = 0; i <= kpo - 1; i++)
        ans = (ans + pp[mi++] * wt[i]) % ((long long)(1e9 + 7));
      cout << ans << "\n";
    }
  }
}
