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 gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powmod(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = powmod(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modif(long long x, long long m) { return (powmod(x, m - 2, m)); }
bool fa(vector<long long> &x, vector<long long> &y) { return x[0] < y[0]; }
bool fa1(vector<long long> &x, vector<long long> &y) { return x[1] < y[1]; }
bool f1(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.second > y.second;
}
bool f2(pair<long long, long long> &x, pair<long long, long long> &y) {
  return x.first > y.first;
}
bool f3(long long &x, long long &y) { return x > y; }
const long long maxn = 1e5 + 10;
vector<vector<long long> > gr;
vector<long long> af;
void dfs(long long s, long long e) {
  for (auto x : gr[s]) {
    if (x != e) dfs(x, s);
  }
  if (e != -1) {
    af[e] += af[s] + 1;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long te;
  cin >> te;
  while (te--) {
    gr.clear();
    af.clear();
    long long n;
    cin >> n;
    vector<long long> temp;
    for (int i = 0; i < n + 10; i++) {
      gr.push_back(temp);
      af.push_back(0ll);
    }
    for (int i = 0; i < n - 1; i++) {
      long long x, y;
      cin >> x >> y;
      x--;
      y--;
      gr[x].push_back(y);
      gr[y].push_back(x);
    }
    long long m;
    cin >> m;
    long long c[m];
    for (int i = 0; i < m; i++) cin >> c[i];
    long long st = -1;
    for (int i = 0; i < n; i++) {
      if (gr[i].size() == 1) {
        st = i;
        break;
      }
    }
    dfs(st, -1);
    vector<long long> h;
    for (int i = 0; i < n; i++) {
      af[i]++;
      if (i == st) continue;
      long long x = af[i];
      long long z = x * (n - x);
      h.push_back(z);
    }
    long long val[n - 1];
    sort(c, c + m);
    for (int i = 0; i < n - 1; i++) val[i] = 1;
    long long re = 1;
    if (m <= n - 1) {
      for (int i = 0; i < m; i++) val[i] = c[i];
    } else {
      for (int i = 0; i < n - 1; i++) val[i] = c[i];
      for (int i = n - 1; i < m; i++) {
        re *= c[i];
        re %= 1000000007;
      }
      val[n - 2] *= re;
      val[n - 2] %= 1000000007;
    }
    sort(val, val + n - 1);
    sort(h.begin(), h.end());
    long long ans = 0;
    for (int i = 0; i < n - 1; i++) {
      h[i] %= 1000000007;
      val[i] %= 1000000007;
      long long x = h[i] * val[i];
      x %= 1000000007;
      ans += x;
      ans %= 1000000007;
      ;
    }
    ans %= 1000000007;
    ;
    cout << ans << endl;
  }
  return 0;
}
