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 inf = 1e18;
const long long mod = 1e9 + 7;
const long long MOD = 998244353;
const long long MAX = 2e5 + 1;
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod)) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return ((a % mod) * (b % mod)) % mod;
}
long long pwr(long long x, long long n) {
  if (n == 0) {
    return 1;
  }
  if (n & 1) {
    return mul(x, pwr(mul(x, x), (n - 1) / 2));
  } else {
    return pwr(mul(x, x), n / 2);
  }
}
long long modInv(long long n) { return pwr(n, mod - 2); }
long long gcd(long long a, long long b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
long long n, m, a[MAX], subtree[MAX];
vector<long long> g[MAX];
long long dfs(long long u = 0, long long par = -1) {
  long long cnt = 1;
  for (auto v : g[u]) {
    if (v != par) {
      cnt += dfs(v, u);
    }
  }
  return subtree[u] = cnt;
}
void speed_solve() {
  cin >> n;
  for (long long i = 0; i < n; i++) g[i].clear();
  vector<pair<long long, long long>> edges;
  for (long long i = 0; i < n - 1; i++) {
    long long u, v;
    cin >> u >> v;
    u--, v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  cin >> m;
  for (long long i = 0; i < m; i++) cin >> a[i];
  sort(a, a + m);
  reverse(a, a + m);
  if (m < n - 1) {
    for (long long i = m; i < n - 1; i++) a[i] = 1;
  } else if (m > n - 1) {
    long long b[n - 1], k = 1;
    b[0] = 1;
    for (long long i = 0; i < m; i++) {
      if (i < m - n + 2) {
        b[0] = mul(b[0], a[i]);
      } else {
        b[k++] = a[i];
      }
    }
    for (long long i = 0; i < n - 1; i++) {
      a[i] = b[i];
    }
  }
  dfs();
  vector<long long> contribution(n - 1);
  for (long long i = 0; i < n - 1; i++) {
    contribution[i] = subtree[i + 1] * (n - subtree[i + 1]);
  }
  sort(contribution.begin(), contribution.end());
  reverse(contribution.begin(), contribution.end());
  long long ans = 0;
  for (long long i = 0; i < n - 1; i++) {
    ans = add(ans, mul(contribution[i], a[i]));
  }
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test_cases = 1;
  cin >> test_cases;
  for (long long i = 0; i < test_cases; i++) {
    speed_solve();
  }
  return 0;
}
