#include <bits/stdc++.h>
using namespace std;
const long long N = 3e5 + 5;
const long long mod = 1e9 + 7;
vector<vector<long long> > al;
vector<long long> sum;
vector<long long> mul;
long long n;
void dfs(long long u, long long p = -1) {
  sum[u] = 1;
  for (long long to : al[u]) {
    if (to == p) continue;
    dfs(to, u);
    sum[u] += sum[to];
  }
  if (u != 1) mul.push_back((sum[u] * (n - sum[u])));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    al.clear();
    al.resize(n + 1);
    sum.clear();
    sum.resize(n + 1, 0);
    mul.clear();
    long long u, v;
    for (long long i = 1; i < n; ++i) {
      cin >> u >> v;
      al[u].push_back(v);
      al[v].push_back(u);
    }
    long long m;
    cin >> m;
    vector<long long> p(m);
    for (long long i = 0; i < m; ++i) {
      cin >> p[i];
    }
    dfs(1);
    sort(p.begin(), p.end());
    sort(mul.begin(), mul.end());
    assert(mul.size() == n - 1);
    long long res = 0;
    if (m <= n - 1) {
      while (p.size() < n - 1) {
        p.push_back(1);
      }
      sort(p.begin(), p.end());
      for (long long i = n - 2; i >= 0; --i) {
        res = (res + p[i] * mul[i]) % mod;
      }
    } else {
      long long j = 0;
      for (long long i = 0; i < n - 2; ++i) {
        res = (res + p[j++] * mul[i]) % mod;
      }
      long long tot = 1;
      while (j < m) {
        tot = (tot * p[j++]) % mod;
      }
      res = (res + tot * mul[n - 2]) % mod;
    }
    cout << (res + mod) % mod << endl;
  }
}

### Prompt

Create a solution in Cpp for 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 N = 3e5 + 5;
const long long mod = 1e9 + 7;
vector<vector<long long> > al;
vector<long long> sum;
vector<long long> mul;
long long n;
void dfs(long long u, long long p = -1) {
  sum[u] = 1;
  for (long long to : al[u]) {
    if (to == p) continue;
    dfs(to, u);
    sum[u] += sum[to];
  }
  if (u != 1) mul.push_back((sum[u] * (n - sum[u])));
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    cin >> n;
    al.clear();
    al.resize(n + 1);
    sum.clear();
    sum.resize(n + 1, 0);
    mul.clear();
    long long u, v;
    for (long long i = 1; i < n; ++i) {
      cin >> u >> v;
      al[u].push_back(v);
      al[v].push_back(u);
    }
    long long m;
    cin >> m;
    vector<long long> p(m);
    for (long long i = 0; i < m; ++i) {
      cin >> p[i];
    }
    dfs(1);
    sort(p.begin(), p.end());
    sort(mul.begin(), mul.end());
    assert(mul.size() == n - 1);
    long long res = 0;
    if (m <= n - 1) {
      while (p.size() < n - 1) {
        p.push_back(1);
      }
      sort(p.begin(), p.end());
      for (long long i = n - 2; i >= 0; --i) {
        res = (res + p[i] * mul[i]) % mod;
      }
    } else {
      long long j = 0;
      for (long long i = 0; i < n - 2; ++i) {
        res = (res + p[j++] * mul[i]) % mod;
      }
      long long tot = 1;
      while (j < m) {
        tot = (tot * p[j++]) % mod;
      }
      res = (res + tot * mul[n - 2]) % mod;
    }
    cout << (res + mod) % mod << endl;
  }
}

```