#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long M = 1e9 + 7;
std::vector<long long> e[N], sz(N);
std::vector<long long> es;
long long n;
void dfs(long long v, long long p) {
  sz[v]++;
  for (long long i : e[v]) {
    if (i != p) {
      dfs(i, v);
      es.push_back((n - sz[i]) * sz[i]);
      sz[v] += sz[i];
    }
  }
}
void solve() {
  long long x, y, m, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) e[i].clear(), sz[i] = 0;
  es.clear();
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, 1);
  cin >> m;
  std::vector<long long> p(m);
  for (long long &i : p) cin >> i;
  while (p.size() < es.size()) {
    p.push_back(1);
  }
  sort((p).begin(), (p).end());
  sort((es).begin(), (es).end());
  for (int i = 0; i < es.size() - 1; i++) {
    ans = (ans + es[i] * p[i] % M) % M;
  }
  long long tot = 1;
  for (int i = es.size() - 1; i < p.size(); i++) {
    tot = tot * p[i] % M;
  }
  cout << (ans + tot * es.back() % M) % M << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
long long M = 1e9 + 7;
std::vector<long long> e[N], sz(N);
std::vector<long long> es;
long long n;
void dfs(long long v, long long p) {
  sz[v]++;
  for (long long i : e[v]) {
    if (i != p) {
      dfs(i, v);
      es.push_back((n - sz[i]) * sz[i]);
      sz[v] += sz[i];
    }
  }
}
void solve() {
  long long x, y, m, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) e[i].clear(), sz[i] = 0;
  es.clear();
  for (int i = 0; i < n - 1; i++) {
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, 1);
  cin >> m;
  std::vector<long long> p(m);
  for (long long &i : p) cin >> i;
  while (p.size() < es.size()) {
    p.push_back(1);
  }
  sort((p).begin(), (p).end());
  sort((es).begin(), (es).end());
  for (int i = 0; i < es.size() - 1; i++) {
    ans = (ans + es[i] * p[i] % M) % M;
  }
  long long tot = 1;
  for (int i = es.size() - 1; i < p.size(); i++) {
    tot = tot * p[i] % M;
  }
  cout << (ans + tot * es.back() % M) % M << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  long long TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

```