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.*;
import java.util.Scanner;

public class T4 {
    static long Mod = 1_000_000_007L;
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        for (int i=0;i<n;i++) {
            mainPrint(s);
        }
    }
    public static void mainPrint(Scanner s){
        int n = s.nextInt();
        List<Integer>[] g = new List[n];
        int x,y;
        for (int i = 0; i < n; i++) {
            g[i] = new ArrayList<>();
        }
        for (int i = 0; i < n - 1; i++) {
            x = s.nextInt();
            y = s.nextInt();
            g[x-1].add(y-1);
            g[y-1].add(x-1);
        }
        boolean[] visited = new boolean[n];
        TreeNode root = new TreeNode(0);
        Queue<TreeNode> q = new LinkedList<>();
        q.offer(root);
        visited[0]=true;
        while(!q.isEmpty()){
            TreeNode t = q.poll();
            for(int i:g[(int)t.val]){
                if(!visited[i]){
                    visited[i]=true;
                    TreeNode add = new TreeNode(i);
                    t.children.add(add);
                    q.offer(add);
                }
            }
        }
        setVal(root);
        long[] edgeVal = new long[n];
        setEdgeVal(root,n,edgeVal,new int[]{0});
        Arrays.sort(edgeVal);
        int m = s.nextInt();
        int[] p = new int[m];
        for (int i=0;i<m;i++){
            p[i]=s.nextInt();
        }
        Arrays.sort(p);
        long ans=0;
        int q1=m-1;
//        System.out.println(Arrays.toString(edgeVal));
//        System.out.println(Arrays.toString(p));

        for(int i=n-1;i>=1;i--){
            long target = 1;
            if(i==n-1 && m>n-1) {
                for(;q1>=n-2;){
                    target *= p[q1--];
                    target %= Mod;
                }
            }else {
                if(q1!=-1) target *= p[q1--];
            }
            ans += target*edgeVal[i];
            ans %=Mod;
        }
        System.out.println((int) ans);
    }
    public static void setVal(TreeNode root){
        int sum = 0;
        for(TreeNode t:root.children) {
            setVal(t);
            sum += t.val;
        }
        root.val = sum+1;
    }
    public static void setEdgeVal(TreeNode root,int n,long[] edgeVal,int[] p){
        for(TreeNode t:root.children) {
            setEdgeVal(t,n,edgeVal,p);
        }
        edgeVal[p[0]++] = root.val*(n-root.val);
    }
    static class TreeNode{
        List<TreeNode> children = new LinkedList<>();
        long val;
        public TreeNode(int val) {
            this.val = val;
        }
    }
}
