#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
vector<int> adj[100005];
long long sub[100005];
long long cnt[100005];
long long p[60005];
void dfs(int n, int par) {
  sub[n] = 1;
  for (auto i : adj[n]) {
    if (i == par) continue;
    dfs(i, n);
    sub[n] += sub[i];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int z = 1; z <= t; z++) {
    long long n;
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p[i];
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
      cnt[i] = (n - sub[i]) * 1ll * sub[i];
    }
    sort(p + 1, p + m + 1, greater<long long>());
    sort(cnt + 1, cnt + n + 1, greater<long long>());
    for (int i = 1; i <= n; i++) {
      cnt[i] %= MOD;
    }
    long long res = 0;
    if (m <= n - 1) {
      for (int i = 1; i <= n - 1; i++) {
        if (i <= m)
          res += p[i] * cnt[i] % MOD;
        else
          res += cnt[i];
        res %= MOD;
      }
    } else {
      long long val = 1;
      for (int i = 1; i <= m - n + 2; i++) {
        val *= p[i];
        val %= MOD;
      }
      p[m - n + 2] = val;
      for (int i = 1; i <= n - 1; i++) {
        res += cnt[i] * p[i + m - n + 1] % MOD;
        res %= MOD;
      }
    }
    cout << res << "\n";
    for (int i = 1; i <= n; i++) {
      adj[i].clear();
      sub[i] = 0;
      cnt[i] = 0;
    }
  }
  return 0;
}

### 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 MOD = 1e9 + 7;
vector<int> adj[100005];
long long sub[100005];
long long cnt[100005];
long long p[60005];
void dfs(int n, int par) {
  sub[n] = 1;
  for (auto i : adj[n]) {
    if (i == par) continue;
    dfs(i, n);
    sub[n] += sub[i];
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  cin >> t;
  for (int z = 1; z <= t; z++) {
    long long n;
    cin >> n;
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    int m;
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p[i];
    }
    dfs(1, 0);
    for (int i = 1; i <= n; i++) {
      cnt[i] = (n - sub[i]) * 1ll * sub[i];
    }
    sort(p + 1, p + m + 1, greater<long long>());
    sort(cnt + 1, cnt + n + 1, greater<long long>());
    for (int i = 1; i <= n; i++) {
      cnt[i] %= MOD;
    }
    long long res = 0;
    if (m <= n - 1) {
      for (int i = 1; i <= n - 1; i++) {
        if (i <= m)
          res += p[i] * cnt[i] % MOD;
        else
          res += cnt[i];
        res %= MOD;
      }
    } else {
      long long val = 1;
      for (int i = 1; i <= m - n + 2; i++) {
        val *= p[i];
        val %= MOD;
      }
      p[m - n + 2] = val;
      for (int i = 1; i <= n - 1; i++) {
        res += cnt[i] * p[i + m - n + 1] % MOD;
        res %= MOD;
      }
    }
    cout << res << "\n";
    for (int i = 1; i <= n; i++) {
      adj[i].clear();
      sub[i] = 0;
      cnt[i] = 0;
    }
  }
  return 0;
}

```