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;
template <typename T>
void read(T &x) {
  x = 0;
  char c = getchar();
  int f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  x *= f;
}
const int N = 4e5 + 7, MO = 1e9 + 7;
int p[N], ne[N], head[N], fa[N], m, n, t, x, y;
long long cnt, pp[N], ans, sz[N], a[N];
void jb(int x, int y) {
  p[++cnt] = y;
  ne[cnt] = head[x];
  head[x] = cnt;
}
void dfs(int x) {
  sz[x] = 1;
  for (int i = head[x]; i; i = ne[i]) {
    if (p[i] == fa[x]) continue;
    fa[p[i]] = x;
    dfs(p[i]);
    sz[x] += sz[p[i]];
  }
  a[x] = sz[x] * (n - sz[x]);
}
int main() {
  read(t);
  for (int k = 1; k <= t; k++) {
    read(n);
    ans = 0;
    cnt = 0;
    memset(head, 0, sizeof(head));
    memset(ne, 0, sizeof(ne));
    memset(p, 0, sizeof(p));
    for (int i = 1; i < n; i++) {
      read(x);
      read(y);
      jb(x, y);
      jb(y, x);
    }
    dfs(1);
    read(m);
    for (int i = 1; i <= m; i++) read(pp[i]);
    sort(pp + 1, pp + m + 1);
    if (m >= n) {
      cnt = 1;
      for (int i = n - 1; i <= m; i++) cnt = (cnt * pp[i]) % MO;
      m = n - 1;
      pp[n - 1] = cnt;
    }
    sort(a + 1, a + n + 1);
    for (int i = 1; i <= n; i++) a[i] = a[i] % MO;
    for (int i = n; i > 1; i--) {
      if (m > 0)
        ans = (ans + pp[m] * a[i] % MO) % MO;
      else
        ans = (ans + a[i]) % MO;
      m--;
    }
    cout << ans << endl;
  }
}
