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 = 1000000000 + 7;
const long long N = 100005;
const long long M = 6 * 10005;
long long k[M];
long long dp[N];
vector<long long> graph[N];
vector<long long> ePaths;
void dfs(long long v, long long p) {
  dp[v] = 1;
  for (auto u : graph[v]) {
    if (u != p) {
      dfs(u, v);
      dp[v] += dp[u];
    }
  }
}
void dfs2(long long v, long long p) {
  for (auto u : graph[v]) {
    if (p == u) continue;
    ePaths.push_back((dp[1] - dp[u]) * dp[u]);
    dfs2(u, v);
  }
}
void prEpa() {
  cout << "\nepa: ";
  for (auto x : ePaths) {
    cout << x << " ";
  }
  cout << "\n";
}
void solve() {
  for (long long i = 0; i < N; ++i) {
    graph[i].clear();
    dp[i] = 0;
  }
  ePaths.clear();
  for (long long i = 0; i < M; ++i) {
    k[i] = 0;
  }
  long long n;
  cin >> n;
  for (long long i = 0; i < n - 1; ++i) {
    long long x, y;
    cin >> x >> y;
    graph[x].push_back(y);
    graph[y].push_back(x);
  }
  long long m;
  cin >> m;
  for (long long i = 0; i < m; ++i) {
    cin >> k[i];
  }
  dfs(1, -1);
  dfs2(1, -1);
  sort(ePaths.begin(), ePaths.end());
  reverse(ePaths.begin(), ePaths.end());
  sort(k, k + m);
  reverse(k, k + m);
  long long res = 0;
  if ((long long)ePaths.size() >= m) {
    for (long long i = 0; i < (long long)ePaths.size(); ++i) {
      if (i < m) {
        res = (res + (k[i] * ePaths[i]) % MOD) % MOD;
      } else {
        res = (res + (ePaths[i]) % MOD) % MOD;
      }
    }
  } else {
    long long fEls = m - ePaths.size();
    long long fRes = 1;
    for (long long i = 0; i <= fEls; ++i) {
      fRes = (fRes * k[i]) % MOD;
    }
    res += (ePaths[0] * fRes) % MOD;
    long long j = 1;
    for (long long i = fEls + 1; i < m; ++i, ++j) {
      res = (res + (ePaths[j] * k[i]) % MOD) % MOD;
    }
  }
  cout << res << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}
