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;
int dfs(int i, vector<vector<int>>& edg, vector<int>& color,
        vector<long long int>& parts) {
  int n = edg[i].size();
  int sum = 0;
  int j;
  for (j = 0; j < n; j++) {
    if (color[edg[i][j]] == -1) {
      color[edg[i][j]] = 0;
      sum += dfs(edg[i][j], edg, color, parts);
    }
  }
  color[i] = 1;
  parts.push_back(sum + 1);
  return sum + 1;
}
int main() {
  int t;
  int n;
  int m;
  int ind1, ind2;
  cin >> t;
  int i, x, y;
  vector<long long int> parts;
  vector<long long int> primes;
  vector<long long int> prime;
  long long int product;
  vector<int> color;
  long long int ans = 0;
  long long int val, val1, val2;
  while (t > 0) {
    cin >> n;
    vector<vector<int>> edg(n + 1);
    parts.clear();
    color.clear();
    prime.clear();
    primes.clear();
    ans = 0;
    for (i = 1; i < n; i++) {
      cin >> x;
      cin >> y;
      edg[x].push_back(y);
      edg[y].push_back(x);
      color.push_back(-1);
    }
    color.push_back(-1);
    color.push_back(-1);
    color[1] = 0;
    dfs(1, edg, color, parts);
    sort(parts.begin(), parts.end());
    ind2 = upper_bound(parts.begin(), parts.end(), n / 2) - parts.begin();
    ind1 = ind2 - 1;
    cin >> m;
    for (i = 0; i < m; i++) {
      cin >> x;
      primes.push_back(x);
    }
    sort(primes.begin(), primes.end(), greater<int>());
    if (m > n - 1) {
      product = 1;
      for (i = 0; i < m - n + 2; i++) {
        product *= primes[i];
        product %= 1000000007;
      }
      prime.push_back(product);
      for (i = m - n + 2; i < m; i++) {
        prime.push_back(primes[i]);
      }
    } else {
      for (i = 0; i < m; i++) {
        prime.push_back(primes[i]);
      }
    }
    ans = 0;
    m = prime.size();
    for (i = 0; i < m; i++) {
      if (ind1 >= 0 && ind2 < n - 1) {
        val1 = parts[ind1];
        val1 *= (n - parts[ind1]);
        val2 = parts[ind2];
        val2 *= (n - parts[ind2]);
        val = max(val1, val2);
        if (val == val1) {
          ind1--;
        } else {
          ind2++;
        }
        val %= (1000000007);
      } else if (ind1 >= 0) {
        val1 = parts[ind1];
        val1 *= (n - parts[ind1]);
        val = val1;
        val %= (1000000007);
        ind1--;
      } else {
        val2 = parts[ind2];
        val2 *= (n - parts[ind2]);
        val = val2;
        val %= (1000000007);
        ind2++;
      }
      val *= prime[i];
      val %= (1000000007);
      ans += val;
      ans %= (1000000007);
    }
    while (ind1 >= 0) {
      val = parts[ind1];
      val *= (n - parts[ind1]);
      val %= (1000000007);
      ans += val;
      ans %= (1000000007);
      ind1--;
    }
    while (ind2 < n - 1) {
      val = parts[ind2];
      val *= (n - parts[ind2]);
      val %= (1000000007);
      ans += val;
      ans %= (1000000007);
      ind2++;
    }
    cout << ans << endl;
    edg.clear();
    t--;
  }
  return 0;
}
