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;
const long long inf = 1e18;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long pwr(long long x, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n & 1) {
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  } else {
    return pwr(mul(x, x), n / 2);
  }
}
long long modInv(long long n) { return pwr(n, mod - 2); }
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
vector<vector<int>> adj;
vector<long long> visi;
vector<long long> vi;
int n;
void dfs(int u, int p) {
  visi[u] = 1;
  for (auto& a : adj[u]) {
    if (a != p) {
      dfs(a, u);
      visi[u] += visi[a];
    }
  }
  if (u != 1) vi.push_back((visi[u] * (n - visi[u])));
}
void solve() {
  cin >> n;
  adj.clear();
  visi.clear();
  vi.clear();
  adj.resize(n + 1);
  visi.resize(n + 1);
  for (long long i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  int m;
  cin >> m;
  vector<int> pf(m);
  for (auto& c : pf) cin >> c;
  sort(pf.begin(), pf.end());
  reverse(pf.begin(), pf.end());
  dfs(1, -1);
  sort(vi.begin(), vi.end());
  reverse(vi.begin(), vi.end());
  vector<long long> ans(n);
  if (m <= n - 1) {
    for (int i = 0; i < n - 1; ++i) {
      if (i < m)
        ans[i] = (vi[i] * pf[i]) % mod;
      else
        ans[i] = 1ll * vi[i] % mod;
    }
  } else {
    int gp = m - n + 1 + 1;
    ans[0] = vi[0];
    for (int i = 0; i < gp; ++i) {
      ans[0] = (ans[0] * pf[i]) % mod;
    }
    for (int i = 1; i < n - 1; ++i) {
      ans[i] = (pf[gp + i - 1] * vi[i]) % mod;
    }
  }
  int ok = 0;
  for (int i = 0; i < n - 1; ++i) {
    ok = (ok + ans[i]);
    ok = ok % mod;
  }
  cout << ok << '\n';
}
int main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}
