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.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;
import java.util.StringTokenizer;

public class D665 {
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    StringTokenizer st = new StringTokenizer(br.readLine());
    int T = Integer.parseInt(st.nextToken());
    for (int t = 0; t < T; t++) {
      // take input
      st = new StringTokenizer(br.readLine());
      int n = Integer.parseInt(st.nextToken());
      List<Integer>[] adjList = new List[n];
      for (int i = 0; i < n; i++) {
        adjList[i] = new ArrayList<>();
      }
      for (int i = 0; i < n - 1; i++) {
        st = new StringTokenizer(br.readLine());
        int x = Integer.parseInt(st.nextToken()) - 1;
        int y = Integer.parseInt(st.nextToken()) - 1;
        adjList[x].add(y);
        adjList[y].add(x);
      }
      st = new StringTokenizer(br.readLine());
      int m = Integer.parseInt(st.nextToken());
      st = new StringTokenizer(br.readLine());
      Long[] factors = new Long[m];
      for (int i = 0; i < m; i++) factors[i] = Long.parseLong(st.nextToken());
      Arrays.sort(factors, Collections.reverseOrder());

      // get the number of times an edge would be traversed.
      numPaths = new ArrayList<>();
      dfs(0, -1, adjList);
      Collections.sort(numPaths, Collections.reverseOrder());

      // multiply factor with numPaths.
      long lastFactor = factors[0];
      for (int i = 1; i <= m - n + 1; i++) {
        lastFactor = (lastFactor * factors[i]) % MOD;
      }
      factors[Math.max(0, m - n + 1)] = lastFactor;
      long sum = 0;
      for (int i = 0; i < n - 1; i++) {
        int idx = Math.max(0, m - n + 1) + i;
        long f = idx < m ? factors[idx] : 1;
        sum = (sum + (f * numPaths.get(i)) % MOD) % MOD;
      }
      System.out.println(sum);
    }
  }

  private static List<Long> numPaths;
  private static final int MOD = 1000000007;

  // calculates the number of nodes in the subtree of x if prev was the parent node of x.
  private static long dfs(int x, int prev, List<Integer>[] adjList) {
    long numNodes = 1; // including the curr.
    for (Integer i : adjList[x]) {
      if (i != prev) numNodes += dfs(i, x, adjList);
    }
    if (numNodes != adjList.length) numPaths.add(numNodes * (adjList.length - numNodes));
    return numNodes;
  }
}
