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;
long long power(long long x, long long y, long long mod) {
  long long res = 1;
  while (y) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
long long binomial(long long n, long long k) {
  long long res = 1;
  if (k > n - k) k = n - k;
  for (long long i = 0; i < k; ++i) {
    res *= (n - i);
    res /= (i + 1);
  }
  return res;
}
const long long mod = 1e9 + 7;
const long long N = 2e5 + 9;
void doper() {}
vector<long long> g[N];
vector<long long> tmp;
long long n;
long long dfs(long long ch, long long par = -1) {
  long long cnt = 1;
  for (auto i : g[ch]) {
    if (i == par) continue;
    cnt += dfs(i, ch);
  }
  if (tmp.size() != n - 1) tmp.push_back(cnt * (n - cnt));
  return cnt;
}
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  doper();
  long long T;
  cin >> T;
  while (T--) {
    tmp.clear();
    for (long long i = 0; i < N; i++) g[i].clear();
    cin >> n;
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    long long m;
    cin >> m;
    vector<long long> p;
    for (long long i = 0; i < m; i++) {
      long long x;
      cin >> x;
      p.push_back(x);
    }
    while (p.size() < n - 1) p.push_back(1);
    sort(p.begin(), p.end());
    while (p.size() > n - 1) {
      long long x = p.back();
      p.pop_back();
      long long y = p.back();
      p.pop_back();
      x = (x * y) % mod;
      p.push_back(x);
    }
    reverse(p.begin(), p.end());
    if (n == 2) tmp.push_back(1);
    for (long long i = 1; i <= n; i++) {
      if (g[i].size() > 1) {
        dfs(i);
        break;
      }
    }
    sort(tmp.rbegin(), tmp.rend());
    long long ans = 0;
    for (long long i = 0; i < n - 1; i++) {
      ans = (ans + (tmp[i] * p[i]) % mod) % mod;
    }
    cout << ans;
    cout << "\n";
    ;
  }
  return 0;
}
