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 INF = 1000000000;
const long long MOD = 1000000007;
vector<long long> egs, son;
vector<long long> G[210000];
long long n;
void dfss(long long p, long long fa = 0) {
  for (auto adj : G[p]) {
    if (adj == fa) continue;
    dfss(adj, p);
    son[p] += son[adj];
  }
}
void reroot(long long p, long long fa = 0) {
  for (auto adj : G[p]) {
    if (adj == fa) continue;
    egs.push_back((son[p] - son[adj]) * son[adj]);
    son[adj] += son[p] - son[adj];
    reroot(adj, p);
  }
}
void solve() {
  cin >> n;
  egs.clear();
  son.assign(n + 1, 1LL);
  for (long long i = 0; i <= n; i++) G[i].clear();
  for (long long i = 0; i < n - 1; i++) {
    long long a, b;
    cin >> a >> b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  dfss(1);
  reroot(1);
  long long np;
  cin >> np;
  vector<long long> snp;
  for (int i = 0; i < np; i++) {
    int tmp;
    cin >> tmp;
    if (tmp == 1) continue;
    snp.push_back(tmp);
  }
  while (snp.size() < egs.size()) snp.push_back(1);
  sort(snp.begin(), snp.end());
  sort(egs.begin(), egs.end());
  long long sp = snp.size() - 1;
  if (snp.size() > egs.size()) {
    long long now = 1;
    while (sp >= egs.size()) {
      now *= snp[sp--];
      now %= MOD;
    }
    snp[sp] *= now;
    snp[sp] %= MOD;
  }
  long long ans = 0;
  for (int i = egs.size() - 1; i >= 0; i--) {
    ans += ((egs[i] % MOD) * snp[sp--]) % MOD;
    ans %= MOD;
  }
  cout << ans % MOD << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  long long t;
  cin >> t;
  for (long long i = 0; i < t; i++) solve();
}
