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;
long long read() {
  long long x = 0, F = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') F = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = x * 10 + c - '0';
    c = getchar();
  }
  return x * F;
}
int ans;
int T, n, m, siz[1200000 + 5], num[1200000 + 5];
vector<long long> P;
vector<int> G[1200000 + 5];
int add(int a, int b) {
  return (a + b >= 1000000007) ? a + b - 1000000007 : a + b;
}
int dec(int a, int b) { return (a - b < 0) ? a - b + 1000000007 : a - b; }
int mul(int a, int b) { return (1LL * a * b) % 1000000007; }
void dfs(int x, int fa) {
  siz[x] = 1;
  for (int i = 0; i < G[x].size(); i++) {
    int v = G[x][i];
    if (v == fa) continue;
    dfs(v, x);
    P.push_back(1LL * siz[v] * (n - siz[v]));
    siz[x] += siz[v];
  }
}
int main() {
  T = read();
  while (T--) {
    P.clear(), ans = 0;
    n = read();
    for (int i = 1; i < n; i++) {
      int u = read(), v = read();
      G[u].push_back(v);
      G[v].push_back(u);
    }
    dfs(1, 0);
    sort(P.begin(), P.end());
    m = read();
    for (int i = 1; i <= m; i++) num[i] = read();
    sort(num + 1, num + m + 1);
    for (int i = m; i > n - 1; i--)
      num[i - 1] = mul(num[i - 1], num[i]), num[i] = 0;
    m = min(m, n - 1);
    reverse(num + 1, num + m + 1);
    for (int i = n - 1; i >= 1; i--)
      ans = add(ans, mul(num[n - i] ? num[n - i] : 1, P[i - 1] % 1000000007));
    printf("%d\n", ans);
    for (int i = 1; i <= n; i++) G[i].clear();
    for (int i = 1; i <= m; i++) num[i] = 0;
  }
}
