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 <int mod>
struct ModInt {
  int x;
  ModInt() : x(0) {}
  ModInt(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}
  ModInt &operator+=(const ModInt &p) {
    if ((x += p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator-=(const ModInt &p) {
    if ((x += mod - p.x) >= mod) x -= mod;
    return *this;
  }
  ModInt &operator*=(const ModInt &p) {
    x = (int)(1LL * x * p.x % mod);
    return *this;
  }
  ModInt &operator/=(const ModInt &p) {
    *this *= p.inv();
    return *this;
  }
  ModInt operator-() const { return ModInt(-x); }
  ModInt operator+(const ModInt &p) const { return ModInt(*this) += p; }
  ModInt operator-(const ModInt &p) const { return ModInt(*this) -= p; }
  ModInt operator*(const ModInt &p) const { return ModInt(*this) *= p; }
  ModInt operator/(const ModInt &p) const { return ModInt(*this) /= p; }
  bool operator==(const ModInt &p) const { return x == p.x; }
  bool operator!=(const ModInt &p) const { return x != p.x; }
  ModInt inv() const {
    int a = x, b = mod, u = 1, v = 0, t;
    while (b > 0) {
      t = a / b;
      swap(a -= t * b, b);
      swap(u -= t * v, v);
    }
    return ModInt(u);
  }
  ModInt pow(int64_t n) const {
    ModInt ret(1), mul(x);
    while (n > 0) {
      if (n & 1) ret *= mul;
      mul *= mul;
      n >>= 1;
    }
    return ret;
  }
  friend ostream &operator<<(ostream &os, const ModInt &p) { return os << p.x; }
  friend istream &operator>>(istream &is, ModInt &a) {
    int64_t t;
    is >> t;
    a = ModInt<mod>(t);
    return (is);
  }
  static int get_mod() { return mod; }
};
const int mod = 1e9 + 7;
using mint = ModInt<mod>;
long long n;
long long dfs(const vector<vector<int>> &adj, vector<long long> &ways, int cur,
              int prev) {
  long long children = 0;
  for (int to : adj[cur]) {
    if (to == prev) continue;
    long long sub = dfs(adj, ways, to, cur);
    children += sub;
    ways.push_back(sub * (n - sub));
  }
  return children + 1;
}
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int t;
  cin >> t;
  while (t--) {
    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);
    }
    long long m;
    cin >> m;
    vector<long long> ps(max(m, n - 1), 1);
    for (int i = 0; i < m; ++i) {
      cin >> ps[i];
    }
    sort(ps.begin(), ps.end());
    vector<mint> pm(ps.size());
    for (int i = 0; i < min((long long)ps.size(), n - 1); ++i) {
      pm[i] = mint(ps[i]);
    }
    for (int i = n - 1; i < m; ++i) {
      pm[n - 2] *= mint(ps[i]);
    }
    vector<long long> ways;
    dfs(adj, ways, 0, -1);
    sort(ways.begin(), ways.end());
    mint ans;
    for (int i = 0; i < n - 1; ++i) {
      ans += pm[i] * mint(ways[i]);
    }
    cout << ans << endl;
  }
}
