#include <bits/stdc++.h>
using namespace std;
const long long M = 1000000007;
long long n;
vector<long long> G[100015];
multiset<long long, greater<long long>> s;
vector<long long> C(100015);
long long dfs(long long u, long long p) {
  C[u] = 0;
  long long tmp = 0;
  for (auto x : G[u]) {
    if (x == p) continue;
    C[x] += dfs(x, u) + 1;
    tmp += C[x];
    s.insert(C[x] * (n - C[x]));
  }
  return tmp;
}
void solve() {
  s.clear();
  cin >> n;
  for (long long i = 1; i < n + 1; ++i) G[i].clear();
  C.clear();
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  long long m;
  cin >> m;
  vector<long long> fac;
  for (long long i = 0; i < m; ++i) {
    long long xx;
    cin >> xx;
    fac.push_back(xx);
  }
  while (fac.size() < n - 1) fac.push_back(1);
  long long ans = 0;
  sort(fac.begin(), fac.end());
  while (fac.size() > n - 1) {
    long long u = fac.back();
    fac.pop_back();
    long long v = fac.back();
    fac.pop_back();
    fac.push_back(((u % M) * (v % M)) % M);
  }
  long long i = fac.size() - 1;
  for (auto x : s) {
    ans += ((x % M) * (fac[i] % M)) % M;
    ans %= M;
    i--;
  }
  cout << ans % M << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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 long long M = 1000000007;
long long n;
vector<long long> G[100015];
multiset<long long, greater<long long>> s;
vector<long long> C(100015);
long long dfs(long long u, long long p) {
  C[u] = 0;
  long long tmp = 0;
  for (auto x : G[u]) {
    if (x == p) continue;
    C[x] += dfs(x, u) + 1;
    tmp += C[x];
    s.insert(C[x] * (n - C[x]));
  }
  return tmp;
}
void solve() {
  s.clear();
  cin >> n;
  for (long long i = 1; i < n + 1; ++i) G[i].clear();
  C.clear();
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  long long m;
  cin >> m;
  vector<long long> fac;
  for (long long i = 0; i < m; ++i) {
    long long xx;
    cin >> xx;
    fac.push_back(xx);
  }
  while (fac.size() < n - 1) fac.push_back(1);
  long long ans = 0;
  sort(fac.begin(), fac.end());
  while (fac.size() > n - 1) {
    long long u = fac.back();
    fac.pop_back();
    long long v = fac.back();
    fac.pop_back();
    fac.push_back(((u % M) * (v % M)) % M);
  }
  long long i = fac.size() - 1;
  for (auto x : s) {
    ans += ((x % M) * (fac[i] % M)) % M;
    ans %= M;
    i--;
  }
  cout << ans % M << '\n';
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```