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.util.*;

public class Solution
{
       
       static void dfs(int n , LinkedList<Integer> arr[] , boolean visited[] , long tot[])
       {
         visited[n] = true;
         for(Integer i : arr[n])
         {
                if(!visited[i])
                {
                     dfs(i , arr , visited , tot);
                     tot[n] += tot[i];
                }
                
               
                
         }
          tot[n]++;
       }
       public static void main(String []args)
       {
              Scanner sc = new Scanner(System.in);
              int t = sc.nextInt();
              while(t-- > 0)
              {
                     int n = sc.nextInt();
                     LinkedList<Integer> arr[] = new LinkedList[n+1];
                     boolean visited[] = new boolean[n+1];
                     for(int i = 1 ; i <= n ; i++)
                     {
                            arr[i] = new LinkedList<Integer>();
                            visited[i] = false;
                     }
                     
                     for(int i = 1 ; i <= n-1 ; i++)
                     {
                            int x = sc.nextInt();
                            int y = sc.nextInt();
                            arr[x].add(y);
                            arr[y].add(x);
                     }
                     long tot[] = new long[n+1];
                     long times[] = new long[n-1];
                     dfs(1,arr,visited,tot);
                     
                     for(int i = 0 ; i < n-1 ; i++)
                     {
                            long nn = n;
                            times[i] = (nn-tot[i+2])*(tot[i+2]);
                     }
                     
                     int m = sc.nextInt();
                     long b[] = new long[m];
                     long mod = 1000000007;
                     for(int i = 0 ; i < m ; i++)
                     {
                            b[i] = sc.nextLong();
                     }
                     Arrays.sort(times);
                     int len = times.length;
                     
                   
                     Arrays.sort(b);
                     long ans = 0;
                     int f = m-1;
                    int tot1 = n-1;
                    int tot2 = m;
                    
                    while(tot2 > tot1)
                    {
                           b[n-2] *= b[f];
                           b[n-2] %= mod;
                           f--;
                           tot2--;
                    }
                    len--;
                   while(len >= 0)
                   {
                          if(f >= 0)
                          {
                                 ans += b[f]*times[len];
                                 ans %= mod;
                                 len--;
                                 f--;
                          }
                          
                          else
                          {
                                 ans += times[len];
                                 ans %= mod;
                                 len--;
                          }
                   }
                   System.out.println(ans);
              }
              
       }
}