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;
using ll = long long;
ll mod = 1e9 + 7;
int t, n, m, p[60005];
ll mul[100001];
vector<int> e[100001];
int dfs(int node, int par) {
  int cnt = 1;
  for (int ch : e[node]) {
    if (ch == par) continue;
    cnt += dfs(ch, node);
  }
  mul[node] = 1ll * cnt * (n - cnt);
  return cnt;
}
void solve() {
  cin >> n;
  for (int i = 0; i < n; ++i) e[i].clear();
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    --x, --y;
    e[x].push_back(y), e[y].push_back(x);
  }
  dfs(0, -1);
  sort(mul + 1, mul + n);
  cin >> m;
  for (int i = 1; i <= m; ++i) cin >> p[i];
  sort(p + 1, p + m + 1);
  if (m <= n - 1) {
    ll ans = 0;
    int j = n - 1;
    for (int i = m; i > 0; --i, --j) {
      ans += 1ll * mul[j] * p[i];
      ans %= mod;
    }
    for (; j > 0; --j) {
      ans += mul[j];
      ans %= mod;
    }
    cout << ans << endl;
  } else {
    ll ans = 0;
    int j = 1;
    for (int i = 1; i + 1 < n; ++i, ++j) {
      ans += 1ll * mul[i] * p[j];
      ans %= mod;
    }
    for (; j <= m; ++j) {
      mul[n - 1] *= p[j];
      mul[n - 1] %= mod;
    }
    cout << (ans + mul[n - 1]) % mod << endl;
  }
}
int main() {
  ios::sync_with_stdio(false);
  for (cin >> t; t--;) solve();
  return 0;
}
