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;
const int MAX = 1e5;
const int mod = 1e9 + 7;
int cnt;
int head[MAX + 5], to[4 * MAX + 5], nex[4 * MAX + 5];
void add(int u, int v) {
  to[++cnt] = v;
  nex[cnt] = head[u];
  head[u] = cnt;
}
int n, m;
int _cnt;
long long prime[MAX + 5], es[MAX + 5];
int dfs(int u, int f) {
  long long sub = 1;
  for (int i = head[u]; i; i = nex[i]) {
    if (to[i] != f) {
      long long temp = dfs(to[i], u);
      es[++_cnt] = temp * (n - temp);
      sub += temp;
    }
  }
  return sub;
}
bool cmp(long long a, long long b) { return a > b; }
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    cnt = 0;
    for (int i = 1; i <= n; i++) head[i] = 0;
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      add(u, v);
      add(v, u);
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) scanf("%lld", prime + i);
    sort(prime + 1, prime + 1 + m);
    if (m >= n) {
      for (int i = n; i <= m; i++) prime[n - 1] = prime[n - 1] * prime[i] % mod;
      m = n - 1;
    }
    _cnt = 0;
    dfs(1, 0);
    sort(es + 1, es + n, cmp);
    long long ans = 0;
    for (int i = 1; i < n; i++) {
      if (m)
        ans = (ans + es[i] * prime[m--] % mod) % mod;
      else
        ans = (ans + es[i]) % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
