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 long long int N = 2e5 + 6, inf = 1e9 + 7;
vector<pair<int, int> > vec[N];
long long int sz[N], n;
pair<long long int, int> P[N];
void dfs(int u, int p) {
  sz[u] = 1;
  for (auto x : vec[u]) {
    int v = x.first;
    if (v == p) continue;
    dfs(v, u);
    sz[u] += sz[v];
    P[x.second] = make_pair(sz[v] * (n - sz[v]), x.second);
  }
}
long long int bigmod(long long int b, long long int p, long long int m) {
  long long int rt = 1;
  while (p) {
    if (p & 1) rt = (rt * b) % m;
    b = (b * b) % m;
    p >>= 1;
  }
  return rt;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    cin >> n;
    for (int i = 1; i <= n; i++) vec[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      vec[u].push_back(make_pair(v, i));
      vec[v].push_back(make_pair(u, i));
    }
    dfs(1, -1);
    sort(P + 1, P + n, greater<pair<long long int, int> >());
    int k;
    cin >> k;
    long long int prim[k + 1];
    for (int i = 1; i <= k; i++) cin >> prim[i];
    sort(prim + 1, prim + k + 1, greater<long long int>());
    long long int rs = 0;
    if (k > n - 1) {
      for (int i = 2; i <= k - n + 2; i++) prim[1] = (prim[1] * prim[i]) % inf;
      for (int i = k - n + 3, j = 2; i <= k; i++) prim[j++] = prim[i];
    }
    for (int i = 1; i < n; i++) {
      rs = (rs + ((i <= k) ? prim[i] * P[i].first : P[i].first)) % inf;
    }
    cout << rs << "\n";
  }
  return 0;
}
