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;
template <class A>
void read(vector<A> &v);
template <class A, size_t S>
void read(array<A, S> &a);
template <class T>
void read(T &x) {
  cin >> x;
}
void read(double &d) {
  string t;
  read(t);
  d = stod(t);
}
void read(long double &d) {
  string t;
  read(t);
  d = stold(t);
}
template <class H, class... T>
void read(H &h, T &...t) {
  read(h);
  read(t...);
}
template <class A>
void read(vector<A> &x) {
  for (auto &a : x) read(a);
}
template <class A, size_t S>
void read(array<A, S> &x) {
  for (auto &a : x) read(a);
}
string to_string(char c) { return string(1, c); }
string to_string(bool b) { return b ? "yes" : "no"; }
string to_string(const char *s) { return string(s); }
string to_string(string s) { return s; }
string to_string(vector<bool> v) {
  string res;
  for (int i = (0); (1) > 0 ? i < ((int)(v).size()) : i > ((int)(v).size());
       i += (1))
    res += char('0' + v[i]);
  return res;
}
template <size_t S>
string to_string(bitset<S> b) {
  string res;
  for (int i = (0); (1) > 0 ? i < (S) : i > (S); i += (1))
    res += char('0' + b[i]);
  return res;
}
template <class T>
string to_string(T v) {
  bool f = 1;
  string res;
  for (auto &x : v) {
    if (!f) res += ' ';
    f = 0;
    res += to_string(x);
  }
  return res;
}
template <class A>
void write(A x) {
  cout << to_string(x) << " ";
}
template <class H, class... T>
void write(const H &h, const T &...t) {
  write(h);
  write(t...);
}
void print() { cout << ("\n"); }
template <class H, class... T>
void print(const H &h, const T &...t) {
  write(h);
  if (sizeof...(t)) write(' ');
  print(t...);
}
void deb() { cerr << endl; }
template <typename Head, typename... Tail>
void deb(Head H, Tail... T) {
  cerr << to_string(H) << " ";
  deb(T...);
}
using ll = long long;
pair<const int, const int> dxdy[4] = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};
const int mod = 1e9 + 7;
void solve() {
  ll n;
  read(n);
  vector<vector<ll>> adj(n + 1);
  for (int i = (0); (1) > 0 ? i < (n - 1) : i > (n - 1); i += (1)) {
    ll x, y;
    read(x, y);
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  vector<ll> node(n + 1, 0);
  function<void(ll, ll)> dfs = [&](ll x, ll p) {
    for (auto i : adj[x]) {
      if (i ^ p) dfs(i, x);
    }
    node[p] += ++node[x];
  };
  dfs(1, 0);
  vector<ll> mul(n - 1);
  for (int i = (0); (1) > 0 ? i < (n - 1) : i > (n - 1); i += (1)) {
    mul[i] = (node[i + 2]) * (n - node[i + 2]);
  }
  sort((mul).begin(), (mul).end(), greater<ll>());
  ll sum = 0;
  for (int i = (0); (1) > 0 ? i < (n - 1) : i > (n - 1); i += (1)) {
    sum += mul[i];
  }
  ll m;
  read(m);
  vector<ll> val(n - 1, 1), in_m(m);
  for (int i = (0); (1) > 0 ? i < (m) : i > (m); i += (1)) cin >> in_m[i];
  sort((in_m).begin(), (in_m).end());
  if (m >= n) {
    for (int i = m - 1; i >= n - 1; i--) {
      in_m[i - 1] *= in_m[i];
      in_m[i - 1] %= mod;
    }
  }
  for (int i = (0); (1) > 0 ? i < (min(n - 1, m)) : i > (min(n - 1, m));
       i += (1)) {
    val[i] = in_m[i];
  }
  sort((val).begin(), (val).end(), greater<ll>());
  ll ans = 0;
  for (int i = (0); (1) > 0 ? i < (n - 1) : i > (n - 1); i += (1)) {
    ll temp = val[i] * mul[i];
    temp %= mod;
    ans += temp;
    ans %= mod;
  }
  write(ans);
  print();
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  cin >> tc;
  for (int i = (0); (1) > 0 ? i < (tc) : i > (tc); i += (1)) {
    solve();
  }
}
