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 int MOD = 1e9 + 7;
int n, m;
long long p[66666], tsf[111111];
long long childcnt[111111];
unordered_map<int, vector<int>> child, adj;
unordered_map<int, bool> chk;
map<pair<int, int>, int> w;
long long cnt[111111];
void reform(int v) {
  chk[v] = true;
  for (int u : adj[v]) {
    if (!chk[u]) {
      child[v].push_back(u);
      reform(u);
    }
  }
}
long long dfs(int v) {
  long long ret = 1;
  for (int u : child[v]) {
    ret += dfs(u);
  }
  return childcnt[v] = ret;
}
void calc(int v) {
  for (int u : child[v]) {
    long long l = childcnt[u], r = childcnt[1] - childcnt[u];
    cnt[u - 1] = l * r;
    calc(u);
  }
}
void solve() {
  cin >> n;
  for (int i = 0; i < n - 1; ++i) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  cin >> m;
  for (int i = 0; i < m; ++i) {
    cin >> p[i];
  }
  sort(p, p + m, greater<long long>());
  for (int i = 0; i < n - 1; ++i) {
    tsf[i] = 1;
  }
  if (m >= n - 1) {
    for (int i = 0; i < m - n + 2; ++i) {
      tsf[0] = (tsf[0] * p[i]) % MOD;
    }
    for (int i = 1; i < n - 1; ++i) {
      tsf[i] = p[i + m - n + 1];
    }
  } else {
    for (int i = 0; i < m; ++i) {
      tsf[i] = p[i];
    }
  }
  reform(1);
  dfs(1);
  calc(1);
  cnt[0] = -1234;
  sort(cnt, cnt + n, greater<long long>());
  long long ans = 0, i = 0;
  for (int i = 0; i < n - 1; ++i) {
    ans += tsf[i] * cnt[i];
    ans %= MOD;
  }
  cout << ans << "\n";
}
void clear() {
  for (int i = 0; i < m; ++i) {
    p[i] = 0;
  }
  for (int i = 0; i < n; ++i) {
    cnt[i] = 0, childcnt[i] = 0, tsf[i] = 0;
  }
  adj.clear();
  child.clear();
  chk.clear();
  w.clear();
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int t;
  cin >> t;
  while (t--) {
    solve();
    clear();
  }
}
