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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
const int mod = 1e9 + 7;
using ll = int64_t;
using namespace std;
template <int MOD_>
struct modnum {
  static constexpr int MOD = MOD_;
  static_assert(MOD_ > 0, "MOD must be positive");
  int v;

 private:
  using ll = long long;
  static int minv(int a, int m) {
    a %= m;
    assert(a);
    return a == 1 ? 1 : int(m - ll(minv(m, a)) * ll(m) / a);
  }

 public:
  modnum() : v(0) {}
  modnum(ll v_) : v(int(v_ % MOD)) {
    if (v < 0) v += MOD;
  }
  explicit operator int() const { return v; }
  friend std::ostream& operator<<(std::ostream& out, const modnum& n) {
    return out << int(n);
  }
  friend std::istream& operator>>(std::istream& in, modnum& n) {
    ll v_;
    in >> v_;
    n = modnum(v_);
    return in;
  }
  friend bool operator==(const modnum& a, const modnum& b) {
    return a.v == b.v;
  }
  friend bool operator!=(const modnum& a, const modnum& b) {
    return a.v != b.v;
  }
  modnum inv() const {
    modnum res;
    res.v = minv(v, MOD);
    return res;
  }
  friend modnum inv(const modnum& m) { return m.inv(); }
  modnum neg() const {
    modnum res;
    res.v = v ? MOD - v : 0;
    return res;
  }
  friend modnum neg(const modnum& m) { return m.neg(); }
  modnum operator-() const { return neg(); }
  modnum operator+() const { return modnum(*this); }
  modnum& operator++() {
    v++;
    if (v == MOD) v = 0;
    return *this;
  }
  modnum& operator--() {
    if (v == 0) v = MOD;
    v--;
    return *this;
  }
  modnum& operator+=(const modnum& o) {
    v += o.v;
    if (v >= MOD) v -= MOD;
    return *this;
  }
  modnum& operator-=(const modnum& o) {
    v -= o.v;
    if (v < 0) v += MOD;
    return *this;
  }
  modnum& operator*=(const modnum& o) {
    v = int(ll(v) * ll(o.v) % MOD);
    return *this;
  }
  modnum& operator/=(const modnum& o) { return *this *= o.inv(); }
  friend modnum operator++(modnum& a, int) {
    modnum r = a;
    ++a;
    return r;
  }
  friend modnum operator--(modnum& a, int) {
    modnum r = a;
    --a;
    return r;
  }
  friend modnum operator+(const modnum& a, const modnum& b) {
    return modnum(a) += b;
  }
  friend modnum operator-(const modnum& a, const modnum& b) {
    return modnum(a) -= b;
  }
  friend modnum operator*(const modnum& a, const modnum& b) {
    return modnum(a) *= b;
  }
  friend modnum operator/(const modnum& a, const modnum& b) {
    return modnum(a) /= b;
  }
};
using mi = modnum<mod>;
void solve() {
  int n, m;
  cin >> n;
  vector<vector<int>> adj(n);
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  cin >> m;
  vector<mi> fac(m);
  for (int i = 0; i < m; i++) {
    cin >> fac[i];
  }
  auto comp = [&](mi X, mi Y) -> bool { return X.v < Y.v; };
  sort(fac.begin(), fac.end(), comp);
  if (int(fac.size()) < n - 1) {
    reverse(fac.begin(), fac.end());
    while (int(fac.size()) < n - 1) {
      if (int(fac.size()) < n - 1) {
        fac.push_back(mi(1));
      }
    }
    reverse(fac.begin(), fac.end());
  }
  if (int(fac.size()) > n - 1) {
    mi cur = 1;
    while (int(fac.size()) >= n - 1) {
      cur *= fac.back();
      fac.pop_back();
    }
    fac.push_back(cur);
  }
  vector<ll> sz(n, 0), num;
  function<void(int, int)> dfs = [&](int node, int par) -> void {
    sz[node] = 1;
    for (auto i : adj[node]) {
      if (i != par) {
        dfs(i, node);
        sz[node] += sz[i];
        num.push_back((ll)sz[i] * (n - sz[i]));
      }
    }
  };
  mi answer = 0;
  dfs(0, 0);
  sort(num.begin(), num.end());
  assert(int(num.size()) >= n - 1);
  assert(int(fac.size()) == n - 1);
  for (int i = 0; i < n - 1; i++) {
    mi cur = (fac[i] * num[i]);
    answer += cur;
  }
  cout << answer << '\n';
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int t(1);
  cin >> t;
  while (t--) {
    solve();
  }
}
