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 N = 100005;
const int MOD = 1000000007;
int T;
int n, m;
struct Edge {
  int to, next;
} edge[N << 1];
int head[N], cnt;
void add_edge(int u, int v) {
  cnt++;
  edge[cnt].to = v;
  edge[cnt].next = head[u];
  head[u] = cnt;
  return;
}
int siz[N];
long long val[N];
void dfs(int u, int father) {
  siz[u] = 1;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v == father) continue;
    dfs(v, u);
    siz[u] += siz[v];
  }
  return;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i < n; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    add_edge(x, y);
    add_edge(y, x);
  }
  dfs(1, 0);
  scanf("%d", &m);
  vector<int> p;
  for (int i = 1; i <= m; i++) {
    int x;
    scanf("%d", &x);
    p.push_back(x);
  }
  sort(p.begin(), p.end());
  while (p.size() > n - 1) {
    int m1 = p[p.size() - 2], m2 = p[p.size() - 1];
    p.pop_back(), p.pop_back();
    p.push_back(1LL * m1 * m2 % MOD);
  }
  reverse(p.begin(), p.end());
  while (p.size() < n - 1) p.push_back(1);
  for (int i = 2; i <= n; i++) val[i] = 1LL * siz[i] * (n - siz[i]);
  sort(val + 2, val + n + 1, greater<long long>());
  long long ans = 0;
  for (int i = 2; i <= n; i++)
    ans = (ans + val[i] % MOD * p[i - 2] % MOD) % MOD;
  printf("%lld\n", ans);
  for (int i = 1; i <= n; i++) head[i] = 0;
  cnt = 0;
  return;
}
int main() {
  scanf("%d", &T);
  while (T--) solve();
  return 0;
}
