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.Arrays;
import java.util.Stack;

public class _1401D {
    BufferedReader bf;
    PrintWriter writer;
    StringBuilder sb;
    static boolean local_system = false;

    long mod = 1_000_000_007L;

    long solve(int[][] e,long[] m){
        int[][] g = packU(e, e.length + 1);
        int[] dp = dfs(g);
        
        long[] count = new long[e.length];
        int i = 0;
        int n = e.length + 1;

        for(int[] ed : e){
            int a = Math.min(dp[ed[0]],dp[ed[1]]);
            int b = n - a;
            count[i++] = ((long)a * (long)b);
        }

        
        Arrays.sort(count);
        Arrays.sort(m);

        int mlen = m.length;
        int clen = count.length;

        long[] mul = new long[n-1];
        Arrays.fill(mul,1L);
        if(mlen < clen){
            for(int ii = mlen - 1,j=n-2;ii>=0;ii--,j--){
                mul[j] = m[ii];
            }
        }else{
            for(int ii = mlen - 1;ii>=0;ii--){
                int j = Math.min(ii,n-2);
                mul[j] = (mul[j] * m[ii])%mod;
            }
        }
        long sum = 0;
        for(int ii = 0;ii<clen;ii++){
            sum = (sum + ((count[ii]%mod) * mul[ii])%mod)%mod;
        }

        return sum;
    }

    int[] dfs(int[][] g){
        int[] dp = new int[g.length];
        Arrays.fill(dp, 1);
        Stack<int[]> s = new Stack<>();
        s.push(new int[]{0,-1,0});
        while(!s.isEmpty()){
            int[] r = s.pop();
            if(r[2]==-1){
                for(int e : g[r[0]]){
                    if(e!=r[1]){
                        dp[r[0]]+=dp[e];
                    }
                }
            }else{
                r[2] = -1;
                s.push(r);
                for(int e : g[r[0]]){
                    if(e!=r[1]){
                        s.push(new int[]{e,r[0],0});
                    }
                }
            }
        } 
        return dp;
    }


    void run() throws IOException {
        int t = i();
        while(t-->0){
            int n = i();
            int[][] q = new int[n-1][];
            for(int i = 0,h=q.length;i<h;i++){
                q[i] = ni();
                q[i][0]--;
                q[i][1]--;
            }
            i();
            long[] m = nl();
            sb.append(solve(q,m)).append("\n");
        }
        writer.println(sb.toString().trim());
    }

    static int[][] packU(int[][] e,int n){
        int[] q = new int[n];
        int[][] g = new int[n][];
        for(int[] ed : e) {
            q[ed[0]]++;
            q[ed[1]]++;
        }
        for(int i = 0;i<n;i++)
            g[i] = new int[q[i]];
        for(int[] ed : e){
            g[ed[0]][--q[ed[0]]] = ed[1];
            g[ed[1]][--q[ed[1]]] = ed[0];
        }
        return g;
    }
    public static void main(String[] args) throws IOException {
        long start_time = System.currentTimeMillis();
        _1401D obj = new _1401D();
        obj.run();
        long end_time = System.currentTimeMillis();
        if (local_system) obj.writer.println("Time : " + (end_time - start_time));
        obj.close();
    }

    public _1401D(){
        writer = new PrintWriter(System.out);
        bf = new BufferedReader(new InputStreamReader(System.in));
        sb = new StringBuilder();
    }

    public int i() throws IOException {
        return Integer.parseInt(bf.readLine());
    }

    public long l() throws IOException {
        return Long.parseLong(bf.readLine());
    }

    public int[] ni() throws IOException {
        String[] data = bf.readLine().split(" ");
        int[] send = new int[data.length];
        for (int i = 0, h = data.length; i < h; i++) send[i] = Integer.parseInt(data[i]);
        return send;
    }

    public long[] nl() throws IOException {
        String[] data = bf.readLine().split(" ");
        long[] send = new long[data.length];
        for (int i = 0, h = data.length; i < h; i++) send[i] = Long.parseLong(data[i]);
        return send;
    }

    public void close() throws IOException{
        writer.flush();
        writer.close();
        bf.close();
    }
}


