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;
int pow(int x, int y) {
  int res = 1;
  while (y) {
    if (y & 1) res *= x;
    y >>= 1;
    x *= x;
  }
  return res;
}
struct Compare {
  constexpr bool operator()(
      pair<long long, pair<long long, long long> > const& a,
      pair<long long, pair<long long, long long> > const& b) const noexcept {
    return a.first < b.first ||
           (a.first == b.first && a.second.first > b.second.first);
  }
};
void prefix_function(string second, long long arr[]) {
  long long border = 0;
  arr[0] = 0;
  for (long long i = 1; i < second.size(); i++) {
    while (border > 0 && second[i] != second[border]) border = arr[border - 1];
    if (second[i] == second[border])
      border++;
    else
      border = 0;
    arr[i] = border;
  }
}
long long mod = 1000000000 + 7;
long long add(long long a, long long b) {
  return (((a % mod) + (b % mod)) % mod);
}
long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod);
}
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = mul(res, a);
    a = mul(a, a);
    b >>= 1;
  }
  return res;
}
long long child[100005];
bool vis1[100005];
vector<long long> adj[100005];
void dfs(long long x) {
  vis1[x] = true;
  child[x] = 1;
  for (long long i = 0; i < adj[x].size(); i++) {
    long long v = adj[x][i];
    if (!vis1[v]) {
      vis1[v] = true;
      dfs(v);
      child[x] += child[v];
    }
  }
}
int main() {
  long long test;
  cin >> test;
  while (test--) {
    long long n;
    cin >> n;
    for (long long i = 0; i <= n; i++) {
      vis1[i] = false;
      child[i] = 0;
      adj[i].clear();
    }
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    dfs(1);
    vector<long long> sura;
    for (long long i = 2; i <= n; i++) {
      sura.push_back(child[i] * (n - child[i]));
    }
    long long q;
    cin >> q;
    vector<long long> prime;
    for (long long i = 0; i < q; i++) {
      long long k;
      cin >> k;
      prime.push_back(k);
    }
    sort(sura.rbegin(), sura.rend());
    sort(prime.rbegin(), prime.rend());
    long long ma = 0;
    while (prime.size() > n - 1) {
      auto x = *(prime.begin());
      prime.erase(prime.begin());
      prime[0] = mul(prime[0], x);
    }
    while (prime.size() < n - 1) {
      prime.push_back(1);
      q++;
    }
    for (long long i = 0; i < prime.size(); i++) {
      ma = add(ma, mul(sura[i], prime[i]));
    }
    cout << ma << endl;
  }
}
