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.
#include <bits/stdc++.h>
using namespace std;
struct node {
  long long x;
  long long y;
  long long ans;
} a[200005];
int cmp(node xx, node yy) { return xx.ans < yy.ans; }
long long b[100005];
vector<int> q[100005];
long long vis[200005];
long long dp[200005];
long long mod = 1e9 + 7;
int dfs(int num) {
  vis[num] = 1;
  int sum = 1;
  for (int i = 0; i < q[num].size(); i++) {
    if (vis[q[num][i]] == 0) {
      sum = sum + dfs(q[num][i]);
    }
  }
  dp[num] = sum;
  return sum;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      vis[i] = 0;
      q[i].clear();
    }
    for (int i = 1; i <= n - 1; i++) {
      scanf("%lld%lld", &a[i].x, &a[i].y);
      q[a[i].x].push_back(a[i].y);
      q[a[i].y].push_back(a[i].x);
    }
    int nh = dfs(1);
    for (int i = 1; i <= n - 1; i++) {
      a[i].ans =
          min(dp[a[i].x], dp[a[i].y]) * (n - min(dp[a[i].x], dp[a[i].y]));
    }
    sort(a + 1, a + n, cmp);
    int m;
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) {
      scanf("%lld", &b[i]);
    }
    if (m < n - 1) {
      for (int i = m + 1; i <= n - 1; i++) {
        b[i] = 1;
      }
      sort(b + 1, b + n);
    } else if (m > n - 1) {
      sort(b + 1, b + m + 1);
      long long nh = 1;
      for (int i = m; i >= n; i--) {
        nh = nh * b[i] % mod;
      }
      b[n - 1] = b[n - 1] * nh % mod;
      sort(b + 1, b + n - 1);
    }
    sort(b + 1, b + n);
    long long sum = 0;
    for (int i = 1; i <= n - 1; i++) {
      sum = (sum + (a[i].ans * b[i] % mod)) % mod;
    }
    printf("%lld\n", sum);
  }
  return 0;
}
