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;
double PI = (acos(-1));
long long md = 1000000007;
long long pw(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = (c * m);
    m = (m * m);
    b /= 2;
  }
  return c;
}
long long pwmd(long long a, long long b) {
  long long c = 1, m = a;
  while (b) {
    if (b & 1) c = ((c * m)) % md;
    m = (m * m) % md;
    b /= 2;
  }
  return c;
}
long long modinv(long long n) { return pwmd(n, md - 2); }
long long min(long long a, long long b) {
  if (a >= b) return b;
  return a;
}
long long nc2(long long n) { return (1ll * n * (n - 1)) / 2; }
long long nsum(long long n) { return (1ll * n * (n + 1)) / 2; }
bool prime(long long a) {
  for (int i = 2; i * i <= a; i++) {
    if (a % i == 0) return false;
  }
  return true;
}
void Sieve(vector<int> &primes) {
  bool IsPrime[100005];
  memset(IsPrime, true, sizeof(IsPrime));
  for (int p = 2; p * p < 100005; p++) {
    if (IsPrime[p] == true) {
      for (int i = p * p; i < 100005; i += p) IsPrime[i] = false;
    }
  }
  for (int p = 2; p < 100005; p++)
    if (IsPrime[p]) primes.push_back(p);
}
vector<char> sml = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                    'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                    's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
vector<char> cap = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                    'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                    'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
vector<vector<long long>> l(100005);
long long vis[100005];
long long v[100005];
void dfs(long long x) {
  vis[x] = 1;
  for (long long z : l[x]) {
    if (vis[z] == 0) {
      dfs(z);
      v[x] += v[z];
    }
  }
}
void solve() {
  long long n;
  cin >> n;
  vector<pair<long long, long long>> edge;
  for (long long i = 0; i < n; i++) l[i + 1].clear();
  for (long long i = 0; i < n - 1; i++) {
    long long x, y;
    cin >> x >> y;
    l[x].push_back(y);
    l[y].push_back(x);
    edge.push_back({x, y});
  }
  for (long long i = 1; i < n + 1; i++) v[i] = 1, vis[i] = 0;
  long long ind;
  for (long long i = 1; i < n + 1; i++) {
    if (l[i].size() == 1) {
      ind = i;
      break;
    }
  }
  vis[ind] = 1;
  for (long long z : l[ind]) {
    dfs(z);
    v[ind] += v[z];
  }
  long long m;
  cin >> m;
  long long p[m];
  for (long long i = 0; i < m; i++) cin >> p[i];
  vector<long long> vec;
  for (long long i = 0; i < n - 1; i++) {
    long long q = min(v[edge[i].first], v[edge[i].second]);
    vec.push_back(q * (n - q));
  }
  sort(vec.begin(), vec.end());
  long long ans = 0;
  sort(p, p + m);
  if (m > n - 1) {
    for (long long i = 0; i < n - 2; i++) {
      ans = (ans + (vec[i] * p[i]) % md) % md;
    }
    long long cur = 1;
    for (long long i = 0; i < m - n + 2; i++) cur = (cur * p[m - 1 - i]) % md;
    ans = (ans + (cur * vec[n - 2]) % md) % md;
  } else {
    for (long long i = 0; i < n - 1 - m; i++) {
      ans = (ans + vec[i]) % md;
    }
    for (long long i = n - 1 - m; i < n - 1; i++)
      ans = (ans + (vec[i] * p[i - n + m + 1]) % md) % md;
  }
  cout << ans % md << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long tests = 1, cnt = 0;
  cin >> tests;
  while (tests--) {
    solve();
  }
  return 0;
}
