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 maxn = 1e5 + 500, mod = 1e9 + 7;
int head[maxn], num_edge;
struct Edge {
  int next, to;
} ed[maxn * 2];
void add(int from, int to) {
  ed[++num_edge].next = head[from];
  ed[num_edge].to = to;
  head[from] = num_edge;
}
int sz[maxn];
struct side {
  int u, v, rd;
  double d;
} sd[maxn];
bool cmp(side a, side b) { return a.d < b.d; }
void dfs(int now, int fa = 0) {
  sz[now] = 1;
  for (int i = head[now]; i; i = ed[i].next) {
    int to = ed[i].to;
    if (to == fa) continue;
    dfs(to, now);
    sz[now] += sz[to];
  }
}
int fac[maxn];
void work() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v), add(v, u);
    sd[i] = {u, v, 0};
  }
  dfs(1);
  for (int i = 1; i < n; i++) {
    int &u = sd[i].u, &v = sd[i].v;
    if (sz[u] < sz[v]) swap(u, v);
    sd[i].d = 1.0 * sz[v] * (n - sz[v]);
    sd[i].rd = (long long)sz[v] * (n - sz[v]) % mod;
  }
  sort(sd + 1, sd + n, cmp);
  int m;
  scanf("%d", &m);
  for (int i = 1; i <= m; i++) scanf("%d", fac + i);
  for (int i = m + 1; i <= n - 1; i++) fac[i] = 1;
  sort(fac + 1, fac + max(m, n - 1) + 1);
  if (m > n - 1)
    for (int i = n; i <= m; i++)
      fac[n - 1] = (long long)fac[n - 1] * fac[i] % mod;
  int ans = 0;
  for (int i = 1; i < n; i++)
    ans += (long long)fac[i] * sd[i].rd % mod, ans %= mod;
  printf("%d\n", ans);
}
void clear() { memset(head, 0, sizeof(head)), num_edge = 0; }
int main() {
  int t;
  scanf("%d", &t);
  while (t--) work(), clear();
  return 0;
}
