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 T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
auto clk = clock();
long long int mod = 1e9 + 7;
const long long inf = 1e17;
const double eps = 1e-6;
const long long int LOGN = 25;
template <class T>
void RV(vector<T> v) {
  for (T& c : v) cin >> c;
}
template <class T>
void RV(vector<vector<T>> v) {
  long long int n = v.size();
  for (vector<T>& c : v) RV(c);
}
template <class T>
vector<T> V(long long int n, T value = 0) {
  vector<T> v(n, value);
  return v;
}
template <class T>
vector<vector<T>> V(long long int n, long long int m, T value = 0) {
  vector<vector<T>> v(n, vector<T>(m, value));
  return v;
}
long long int pow_mod(long long int a, long long int b, long long int m = mod) {
  long long res = 1;
  while (b) {
    if (b & 1) res = ((long long)res * a) % m;
    a = ((long long)a * a) % m;
    b >>= 1;
  }
  return res;
}
template <typename T>
T gcd(T a, T b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
template <typename T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
template <typename T, typename U>
T add(T a, U b) {
  a = (long long)a + b;
  a = (a + mod) % mod;
  return a;
}
template <typename T, typename U>
T mul(T a, U b) {
  a = ((long long)a * b) % mod;
  return a;
}
template <typename T>
T add(vector<T> v) {
  long long sum = 0;
  for (T x : v) sum = (sum + x) % mod;
  return (T)sum;
}
template <typename T>
T mul(vector<T> v) {
  long long sum = 1LL;
  for (T x : v) sum = (sum * x) % mod;
  return (T)sum;
}
template <typename T>
string to_bin(T num) {
  string binary = "";
  while (num) {
    binary += (num % 2 == 1 ? "1" : "0");
    num >>= 1;
  }
  reverse(binary.begin(), binary.end());
  return binary;
}
vector<vector<long long int>> G;
vector<pair<long long int, long long int>> E;
vector<long long int> primes, in, np;
long long int n;
long long int m;
void dfs(long long int s, long long int par) {
  in[s] = 1;
  for (long long int c : G[s]) {
    if (c == par) continue;
    dfs(c, s);
    in[s] += in[c];
  }
}
void __Solve__() {
  cin >> n;
  G = vector<vector<long long int>>(n + 1);
  E.clear();
  primes.clear();
  in = vector<long long int>(n + 1, 0);
  for (long long int i = 0; i < n - 1; i++) {
    long long int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
    E.push_back({u, v});
  }
  long long int m;
  cin >> m;
  for (long long int i = 0; i < m; i++) {
    long long int p;
    cin >> p;
    primes.push_back(p);
  }
  sort(primes.begin(), primes.end());
  reverse(primes.begin(), primes.end());
  np = vector<long long int>();
  if (m <= n - 1) {
    np = primes;
  } else {
    long long int beg = 1;
    long long int diff = m - n + 2;
    for (long long int i = 0; i < diff; i++) {
      beg *= primes[i];
      beg %= mod;
    }
    np.push_back(beg);
    for (long long int i = diff; i < m; i++) {
      np.push_back(primes[i]);
    }
  }
  dfs(1, 0);
  vector<long long int> prods;
  for (long long int i = 0; i < n - 1; i++) {
    long long int x = E[i].first;
    long long int y = E[i].second;
    if (in[x] > in[y]) swap(x, y);
    prods.push_back(in[x] * (n - in[x]));
  }
  sort(prods.begin(), prods.end());
  reverse(prods.begin(), prods.end());
  long long int fans = 0;
  for (long long int i = 0; i < n - 1; i++) {
    long long int x = (i >= np.size() ? 1 : np[i]);
    long long int y = prods[i] % mod;
    fans += (x * y % mod);
    fans %= mod;
  }
  cout << fans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int test_case = 1;
  cin >> test_case;
  while (test_case--) {
    __Solve__();
  }
}
