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 mod = 1e9 + 7;
long long cl(long double a) {
  if (a > (long long)a) {
    return (long long)a + 1;
  } else {
    return (long long)a;
  }
}
long long flr(long double a) {
  if (a < 0.0) {
    return (long long)a - 1;
  }
  return (long long)a;
}
long long n, m, x, y;
const long long M = 1e5 + 100;
vector<long long> gr[M];
bool vis[M] = {false};
long long p[M];
vector<long long> op;
long long dp[M] = {0};
void dfs(long long cur) {
  vis[cur] = true;
  dp[cur] = 1;
  for (auto x : gr[cur]) {
    if (!vis[x]) {
      dfs(x);
      dp[cur] += dp[x];
      op.push_back(dp[x] * (n - dp[x]));
    }
  }
}
void solve() {
  cin >> n;
  op.clear();
  for (long long i = 0; i < n + 1; i++) {
    gr[i].clear();
    vis[i] = false;
    dp[i] = 0;
  }
  for (long long i = 0; i < n - 1; i++) {
    cin >> x;
    cin >> y;
    gr[x].push_back(y);
    gr[y].push_back(x);
  }
  dfs(1);
  cin >> m;
  for (long long i = 0; i < m; i++) cin >> p[i];
  sort(p, p + m, greater<long long>());
  sort(op.begin(), op.end(), greater<long long>());
  long long ans = 0;
  long long j = 0;
  long long i = 0;
  if (m > n - 1) {
    long long cur = 1;
    while (m - j > n - 2) {
      cur *= p[j];
      cur %= mod;
      j++;
    }
    ans += cur * op[0];
    ans %= mod;
    i++;
    while (i < op.size()) {
      ans += p[j] * op[i];
      ans %= mod;
      i++;
      j++;
    }
    cout << ans << "\n";
  } else {
    long long i = 0;
    while (i < op.size()) {
      if (i < m)
        ans += p[i] * op[i];
      else
        ans += op[i];
      ans %= mod;
      i++;
    }
    cout << ans << "\n";
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) solve();
}
