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.*;
import java.util.*;
public class Ishu
{
    static final int MOD = (int)1e9 + 7;
    static int in1 = 0;
    static List<Integer>[] graph = new ArrayList[1];
    static long[] freq = new long[1];
    static int[] size = new int[1];
    static boolean[] visited = new boolean[1];
    static Scanner scan = new Scanner(System.in);
    static int dfs(int ver, int tot)
    {
    visited[ver] = true;
    int ans = 1;
    int n = graph[ver].size();
    int i;
    for(i=0;i<n;++i)
        {
        int u = graph[ver].get(i) - 1;
        if(!visited[u])
            ans += dfs(u,tot);
        else
            ans += size[u];
        }
      
  //    in1++;  
    if(ans != tot)
        freq[in1++] = ((long)ans) * (tot - ans);    
    size[ver] = ans;
    return ans;
    }
    static void tc()
    {
    int n = scan.nextInt();
    
    in1 = 0;
    graph = new ArrayList[n];
    freq = new long[n-1];
    size = new int[n];
    visited = new boolean[n];
    
    int i,j;
    for(i=0;i<n;++i)
        graph[i] = new ArrayList<Integer>();
        
    for(i=0;i<n-1;++i)
        {
        int u = scan.nextInt();
        int v = scan.nextInt();
        graph[u-1].add(v);
        graph[v-1].add(u);
        }
        
    int m = scan.nextInt();
    int[] p = new int[m];
    for(i=0;i<m;++i)
        p[i] = scan.nextInt();
        
    Arrays.sort(p, 0, m);
    long[] one = new long[n-1];
    if(m <= n - 1)
        {
        j = n - 2;
        for(i=m-1;i>=0;--i)
            one[j--] = p[i];
        while(j >= 0)
            one[j--] = 1;
        }
    else
        {
        for(i=0;i<n-1;++i)
            one[i] = p[i];
            
        while(i < m)
            one[n-2] = (one[n-2] * p[i++]) % MOD;
        }
     
     int tot = dfs(0,n);   
    /*for(i=0;i<n-1;++i)    
        System.out.print(freq[i] + " ");*/
  //  System.out.println();
    Arrays.sort(freq,0,n-1);
    long ans = 0l;
    for(i=0;i<n-1;++i)
        ans = (ans + (one[i] * freq[i]) % MOD) % MOD;
        
    System.out.println(ans);
   // System.out.println("in1 : " + in1);
    }
    public static void main(String[] args)
    {
    int t = 1;
    t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}
