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>
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,abm,mmx,avx,avx2,popcnt,fma")
#pragma GCC optimize("fast-math")
using namespace std;
const long long mod = 1e9 + 7, mod2 = 998244353;
const long double eps = 1e-6;
const long long inf = 1e18;
const int base = 1e9;
const long double PI = 3.14159265358979;
int n, m, t;
vector<vector<int>> g;
vector<long long> edges;
long long p[100 * 1000];
unsigned int bitcount(unsigned long long val) {
  return __builtin_popcount(val) + __builtin_popcount(val >> 32);
}
void local() { freopen("test.in", "r", stdin); }
void remote() {}
void read() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  cin >> t;
}
int dfs(int v, int p = -1) {
  if (g[v].size() == 1 && p != -1) return 1;
  int sz = 1;
  for (const int &e : g[v]) {
    if (e != p) {
      int val = dfs(e, v);
      edges.push_back(1ll * val * (n - val));
      sz += val;
    }
  }
  return sz;
}
int main() {
  remote();
  read();
  while (t--) {
    cin >> n;
    g.clear();
    edges.clear();
    g.resize(n);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    cin >> m;
    for (int i = 0; i < m; i++) cin >> p[i];
    for (int i = m; i < n - 1; i++) p[i] = 1;
    sort(p, p + m, greater<int>());
    dfs(0);
    sort(edges.begin(), edges.end(), greater<long long>());
    int start = 0;
    while (m > n - 1) {
      start++;
      p[start] *= p[start - 1], p[start] %= mod;
      m--;
    }
    long long res = 0;
    for (int i = 0; i < n - 1; i++) {
      res += edges[i] * p[start + i];
      res %= mod;
    }
    cout << res << '\n';
    ;
  }
}
