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.lang.*;
import java.io.*;

public class Solution{
  static long[] replaysMap;
  static HashMap<Integer,ArrayList<Integer>> linkList;
  static int n;
  public static void main(String[] args) throws IOException{
    StringBuilder print = new StringBuilder();
    BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
    long t = Long.parseLong(input.readLine());
    long mod = 1000000007;
    while(t-- > 0){
      n = Integer.parseInt(input.readLine());
      replaysMap = new long[n+1];
      linkList = new HashMap<Integer,ArrayList<Integer>>();
      StringTokenizer st;
      for(int i = 0; i < n-1; i++){
        st = new StringTokenizer(input.readLine());
        int u = Integer.parseInt(st.nextToken());
        int v = Integer.parseInt(st.nextToken());
        if(!linkList.containsKey(u)){
          linkList.put(u,new ArrayList<Integer>());
        }
        if(!linkList.containsKey(v)){
          linkList.put(v,new ArrayList<Integer>());
        }
        linkList.get(u).add(v);
        linkList.get(v).add(u);
      }

      int m = Integer.parseInt(input.readLine());
      st = new StringTokenizer(input.readLine());
      long[] pList = new long[Math.max(m,n)];
      for(int i = 0; i < Math.max(m,n); i++){
        long p = 1;
        if(i < m){
          p = Long.parseLong(st.nextToken());
        }
        pList[i] = p;
      }
      long totalSum = 0;
      findReplays(1,0);
      replaysMap[0] = 0;
      replaysMap[1] = 0;
      Arrays.sort(replaysMap);
      Arrays.sort(pList);
      if(m > n-1){
        totalSum = replaysMap[n];
        for(int i = 0; i <= m - (n-1); i++){
          totalSum = (totalSum * pList[Math.max(m,n)- i - 1]) % mod;
          //System.out.printf("i: %d, totalSum: %d, replaysMap[]: %d, pList[]: %d%n", i, totalSum, replaysMap[n], pList[Math.max(m,n) - i - 1]);

        }
        for(int i = m + 2 - n; i < m; i++){

          totalSum = (totalSum + replaysMap[m + 1 -i] * pList[Math.max(m,n) - i - 1]) % mod;
          //System.out.printf("i: %d, totalSum: %d, replaysMap[]: %d, pList[]: %d%n", i, totalSum, replaysMap[m-i], pList[Math.max(m,n) - i - 1]);

        }
      }else{
        for(int i = 0; i < n-1; i++){
          totalSum = (totalSum + replaysMap[n - i] * pList[Math.max(m,n) - i - 1]) % mod;
          //System.out.printf("i: %d, totalSum: %d, replaysMap[]: %d, pList[]: %d%n", i, totalSum, replaysMap[n - i - 1], pList[Math.max(m,n) - i - 1]);
        }
      }
      print.append(totalSum);
      print.append("\n");
    }
    System.out.println(print.toString());
  }

  private static long findReplays(int cur, int prev){
    long replays = 1;
    for(int next: linkList.get(cur)){
      if(next != prev){
        replays += findReplays(next, cur);
      }
    }
    long replaysTotal = replays * (n - replays);
    //sSystem.out.printf("cur: %d, prev: %d, replays: %d, replaysTotal: %d%n", cur, prev, replays, replaysTotal);

    replaysMap[cur] = replaysTotal;
    return replays;
  }
}
