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;
using ll = long long;
using ii = pair<ll, ll>;
const int N = 1e5 + 10;
const int MOD = 1e9 + 7;
int n, m;
ll p[N];
vector<int> g[N];
struct edge {
  int u, v;
  ll mul;
} e[N];
ll sub[N];
int up[N];
int subdfs(int u, int f, int id) {
  sub[u] = 1;
  up[u] = id;
  for (auto i : g[u]) {
    int v = e[i].u ^ e[i].v ^ u;
    if (v != f) sub[u] += subdfs(v, u, i);
  }
  return sub[u];
}
int main() {
  int T;
  scanf("%d", &T);
  while (T--) {
    scanf("%d", &n);
    for (int i = 1; i < n; i++) {
      scanf("%d %d", &e[i].u, &e[i].v);
      g[e[i].u].push_back(i);
      g[e[i].v].push_back(i);
    }
    subdfs(1, 0, 0);
    for (int i = 2; i <= n; i++) {
      int id = up[i];
      e[id].mul = sub[i] * (n - sub[i]);
    }
    sort(e + 1, e + n, [](edge x, edge y) { return x.mul > y.mul; });
    scanf("%d", &m);
    for (int i = 0; i < m; i++) {
      scanf("%lld", &p[i]);
    }
    sort(p, p + m);
    int sz = m;
    while (sz > n - 1) {
      p[sz - 2] = p[sz - 2] * p[sz - 1] % MOD;
      sz--;
    }
    ll ans = 0;
    for (int i = 1; i < n; i++) {
      ll y = (sz ? p[--sz] : 1);
      ans += e[i].mul * y % MOD;
      if (ans >= MOD) ans -= MOD;
    }
    printf("%lld\n", ans);
    for (int i = 1; i <= n; i++) g[i].clear();
  }
  return 0;
}
