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 = 300010;
const long long Mod = 1000000007;
vector<int> g[maxn];
long long siz[maxn];
long long a[maxn];
long long b[maxn];
int cnt;
int n;
void dfs(int u, int fa) {
  siz[u] = 1;
  for (int v : g[u]) {
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
  }
  for (int v : g[u]) {
    if (v == fa) continue;
    a[++cnt] = 1ll * siz[v] * (n - siz[v]);
  }
  return;
}
bool cmp(long long a, long long b) { return a > b; }
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(0);
  ;
  int t;
  cin >> t;
  while (t--) {
    cin >> n;
    int u, v;
    for (int i = 0; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    cnt = 0;
    dfs(1, 0);
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> b[i];
    }
    sort(a + 1, a + cnt + 1, cmp);
    sort(b + 1, b + m + 1, cmp);
    for (int i = 1; i <= cnt; i++) {
      a[i] = a[i] % Mod;
    }
    long long ans = 0;
    if (m > cnt) {
      long long d = 1ll;
      long long pos = m - cnt + 1;
      for (int i = 1; i <= pos; i++) {
        d = 1ll * d * b[i] % Mod;
      }
      ans = (1ll * ans + 1ll * a[1] * d % Mod) % Mod;
      for (int i = 2; i <= cnt; i++) {
        ans = (1ll * ans + 1ll * a[i] * b[i + pos - 1] % Mod) % Mod;
      }
    } else {
      for (int i = 1; i <= cnt; i++) {
        if (i > m) {
          ans = (1ll * ans + 1ll * a[i] % Mod) % Mod;
        } else {
          ans = (1ll * ans + 1ll * a[i] * b[i] % Mod) % Mod;
        }
      }
    }
    cout << (ans % Mod + Mod) % Mod << endl;
  }
  return 0;
}
