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.*;

public class MaxDistributionTree {
    public static void main(String[] args) {
        long timeStart = System.nanoTime();
        Utility.FastReader fr = new Utility.FastReader();
        Solver solver = new Solver(fr);
        int t = fr.nextInt();
        while (t-- > 0) {
            solver.solve();
        }
        long timeEnd = System.nanoTime();
//        System.out.println("Completed in :" + ((double) (timeEnd - timeStart) / 1000000) + " miliseconds");
    }

    static class Solver {
        Utility.FastReader fr;
        static ArrayList<Integer>[] adj;
        static int[] subtreeSize;
        static int MOD=(int)1e9 + 7;


        public Solver(Utility.FastReader fr) {
            this.fr = fr;
        }

        void dfs(int node, int parent){
            ArrayList<Integer> neighbors=adj[node];
            int size=1;
            for(int nei: neighbors){
                if(nei != parent){
                    dfs(nei, node);
                    size += subtreeSize[nei];
                }
            }
            subtreeSize[node]=size;
        }

        long mul(long a, long b){
            return a * b % MOD;
        }
        void solve() {
            int n=fr.nextInt();
            adj=new ArrayList[n];
            for(int i=0; i<n; ++i) adj[i]=new ArrayList<>();
            subtreeSize=new int[n];
            for(int i=0; i<n-1; ++i){
                int u, v;
                u=fr.nextInt()-1;
                v=fr.nextInt()-1;
                adj[u].add(v);
                adj[v].add(u);
            }
            int m=fr.nextInt();
            int[] primes=new int[m];
            for(int i=0; i<m; ++i){
                primes[i]=fr.nextInt();
            }
            Utility.sort(primes);
//            System.out.println(Arrays.toString(primes));
            long[] edgeWeights=new long[n-1];
            Arrays.fill(edgeWeights, 1);
            if(m>n-1){
                int extra=m-(n-1);
                int i=0;
                while(i<m){
                    if(i<=extra){
                        edgeWeights[0]=mul(edgeWeights[0], primes[m-i-1]);
                    }
                    else{
                        edgeWeights[i-extra]=primes[m-i-1];
                    }
                   i++;
                }
            }
            else{
                for(int i=0; i<m; ++i){
                    edgeWeights[i]=primes[i];
                }
            }
//            System.out.println("edge w" + Arrays.toString(edgeWeights));
            Utility.sort(edgeWeights);
            dfs(0, -1);
            long[] contribution=new long[n-1];
            for(int i=1; i<n; ++i){
                long res=subtreeSize[i]*((long)(n-subtreeSize[i]));
                contribution[i-1]=res;
            }
            Arrays.sort(contribution);
            long ans=0;
            for(int i=0; i<n-1; ++i){
                ans=(ans + mul(edgeWeights[i], contribution[i])) % MOD;
            }
            System.out.println(ans);

        }
    }

    static class Utility {
        static void sort(int[] a) {
            ArrayList<Integer> list = new ArrayList<>();
            for (int i : a) list.add(i);
            Collections.sort(list);
            for (int i = 0; i < a.length; i++) a[i] = list.get(i);
        }
        static void sort(long[] a) {
            ArrayList<Long> list = new ArrayList<>();
            for (long i : a) list.add(i);
            Collections.sort(list);
            for (int i = 0; i < a.length; i++) a[i] = list.get(i);
        }

        static void shuffleArray(long[] arr) {
            int n = arr.length;
            Random rnd = new Random();
            for (int i = 0; i < n; ++i) {
                long tmp = arr[i];
                int randomPos = i + rnd.nextInt(n - i);
                arr[i] = arr[randomPos];
                arr[randomPos] = tmp;
            }
        }

        static void shuffleArray(int[] arr) {
            int n = arr.length;
            Random rnd = new Random();
            for (int i = 0; i < n; ++i) {
                int tmp = arr[i];
                int randomPos = i + rnd.nextInt(n - i);
                arr[i] = arr[randomPos];
                arr[randomPos] = tmp;
            }
        }

        static int gcd(int a, int b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }

        static long gcd(long a, long b) {
            if (b == 0)
                return a;
            return gcd(b, a % b);
        }

        static class FastReader {
            BufferedReader br;
            StringTokenizer st;

            public FastReader() {
                br = new BufferedReader(new
                        InputStreamReader(System.in));
            }

            String next() {
                while (st == null || !st.hasMoreElements()) {
                    try {
                        st = new StringTokenizer(br.readLine());
                    } catch (IOException e) {
                        e.printStackTrace();
                    }
                }
                return st.nextToken();
            }

            int nextInt() {
                return Integer.parseInt(next());
            }

            long nextLong() {
                return Long.parseLong(next());
            }

            double nextDouble() {
                return Double.parseDouble(next());
            }

            String nextLine() {
                String str = "";
                try {
                    str = br.readLine();
                } catch (IOException e) {
                    e.printStackTrace();
                }
                return str;
            }
        }
    }
}

