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.stream.*;
import java.io.*;
import java.math.*;

public class Main { 
    
    static boolean FROM_FILE = false;
    
    static boolean LOCAL = System.getProperty("ONLINE_JUDGE") == null;

    static class FastReader { 
        BufferedReader br; 
        StringTokenizer st; 
        public FastReader() { 
            if (FROM_FILE) {
                try {
                    br = new BufferedReader(new FileReader("input.txt")); 
                } catch (IOException error) {
                }   
            } else {
                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; 
        } 
    }
    static int max(int... nums) {
        int res = Integer.MIN_VALUE;
        for (int num: nums) res = Math.max(res, num);
        return res;
    }
    static int min(int... nums) {
        int res = Integer.MAX_VALUE;
        for (int num: nums) res = Math.min(res, num);
        return res;
    }
    static long max(long... nums) {
        long res = Long.MIN_VALUE;
        for (long num: nums) res = Math.max(res, num);
        return res;
    }
    static long min(long... nums) {
        long res = Long.MAX_VALUE;
        for (long num: nums) res = Math.min(res, num);
        return res;
    }
    
    static FastReader fr = new FastReader();
    static PrintWriter out;
    
    public static void main(String[] args) { 
        if (FROM_FILE) {
            try {
                out = new PrintWriter(new FileWriter("output.txt"));
            } catch (IOException error) {
                
            }
        } else {
            out = new PrintWriter(new OutputStreamWriter(System.out));
        }
        new Main().run();
        out.flush();
        out.close();
    } 
    
    int n, m, k;
    int[] head, next, to;
    int[] p;
    
    void addEdge(int u, int v) {
        to[m] = v;
        next[m] = head[u];
        head[u] = m++;
    }
    
    int[] counts;
    
    int dfs(int e, int from) {
        int v = to[e], cnt = 1;
        for (int i = head[v]; i != -1; i = next[i]) {
            int w = to[i]; if (w == from) continue;
            cnt += dfs(i, v);
        }
        counts[e] = cnt;
        return cnt;
    }
    
    void run() {
        int t = fr.nextInt();
        int mod = 1000000007;
        while (t-- > 0) {
            n = fr.nextInt(); m = 0;
            head = new int[n + 1]; Arrays.fill(head, -1);
            next = new int[2 * (n - 1)];
            to = new int[2 * (n - 1)];
            for (int i = 0; i < n - 1; i += 1) {
                int a = fr.nextInt(), b = fr.nextInt();
                addEdge(a, b);
                addEdge(b, a);
            }
            k = fr.nextInt(); p = new int[k];
            for (int i = 0; i < k; i += 1) p[i] = fr.nextInt();
            counts = new int[m];
            for (int i = head[1]; i != -1; i = next[i]) {
                dfs(i, 1);
            }
            if (LOCAL) {
                out.println(Arrays.toString(to));
                out.println(Arrays.toString(counts));
            }
            
            long[] weights = new long[m / 2]; 
            for (int i = 0; i < n - 1; i += 1) {
                int j = 2 * i;
                if (counts[j] == 0) j ^= 1;
                weights[i] = (long)counts[j] * (n - counts[j]);
            }
            if (LOCAL) {
                out.println(Arrays.toString(weights));
            }
            
            Arrays.sort(weights);
            Arrays.sort(p);
            
            long[] factors = new long[m / 2]; 
            int last = m / 2 - 1;
            if (k >= m / 2) {
                factors[last] = 1L;
                for (int i = last; i < k; i += 1) factors[last] = factors[last] * p[i] % mod;
                for (int i = last - 1; i >= 0; i -= 1) factors[i] = p[i];
            } else {
                for (int i = last; i > last - k; i -= 1) factors[i] = p[k - (m / 2 - i)];
                for (int i = last - k; i >= 0; i -= 1) factors[i] = 1L;
            }
            if (LOCAL) {
                out.println(Arrays.toString(factors));
            }
            
            long res = 0;
            for (int i = 0; i < m / 2; i += 1) {
                res = (res + factors[i] * weights[i]) % mod;
            }
            
            out.println(res);
            
        }
        
    }
}