#include <bits/stdc++.h>
using namespace std;
std::vector<long long> tree[200005];
long long childs[200005];
std::vector<long long> contr;
long long n;
void dfs(long long u, long long par = -1) {
  if (tree[u].size() == 1 && u != 1) {
    childs[u] = 1;
    return;
  }
  for (auto ch : tree[u]) {
    if (ch != par) {
      dfs(ch, u);
      childs[u] += childs[ch];
      contr.push_back(childs[ch] * (n - childs[ch]));
    }
  }
  childs[u]++;
}
void solve() {
  contr.clear();
  cin >> n;
  for (long long i = 0; i <= n; ++i) {
    childs[i] = 0;
    tree[i].clear();
  }
  for (long long i = 0; i < n - 1; ++i) {
    long long x, y;
    cin >> x >> y;
    tree[x].push_back(y);
    tree[y].push_back(x);
  }
  long long m;
  cin >> m;
  long long p[m];
  for (long long i = 0; i < m; ++i) cin >> p[i];
  sort(p, p + m);
  dfs(1);
  long long ans = 0;
  sort(contr.begin(), contr.end());
  if (m <= n - 1) {
    for (long long i = 0; i < n - 1 - m; ++i) {
      ans = (ans + contr[i]) % 1000000007;
    }
    for (long long i = n - 1 - m; i < n - 1; ++i) {
      ans = (ans + (contr[i] * p[i - (n - 1 - m)]) % 1000000007) % 1000000007;
    }
  } else {
    for (long long i = 0; i < n - 2; ++i) {
      ans = (ans + (contr[i] * p[i]) % 1000000007) % 1000000007;
    }
    long long x = 1;
    long long i = n - 2;
    while (i < m) {
      x = (x * p[i]) % 1000000007;
      i++;
    }
    ans = (ans + (contr[n - 2] * x) % 1000000007) % 1000000007;
  }
  cout << ans << endl;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
std::vector<long long> tree[200005];
long long childs[200005];
std::vector<long long> contr;
long long n;
void dfs(long long u, long long par = -1) {
  if (tree[u].size() == 1 && u != 1) {
    childs[u] = 1;
    return;
  }
  for (auto ch : tree[u]) {
    if (ch != par) {
      dfs(ch, u);
      childs[u] += childs[ch];
      contr.push_back(childs[ch] * (n - childs[ch]));
    }
  }
  childs[u]++;
}
void solve() {
  contr.clear();
  cin >> n;
  for (long long i = 0; i <= n; ++i) {
    childs[i] = 0;
    tree[i].clear();
  }
  for (long long i = 0; i < n - 1; ++i) {
    long long x, y;
    cin >> x >> y;
    tree[x].push_back(y);
    tree[y].push_back(x);
  }
  long long m;
  cin >> m;
  long long p[m];
  for (long long i = 0; i < m; ++i) cin >> p[i];
  sort(p, p + m);
  dfs(1);
  long long ans = 0;
  sort(contr.begin(), contr.end());
  if (m <= n - 1) {
    for (long long i = 0; i < n - 1 - m; ++i) {
      ans = (ans + contr[i]) % 1000000007;
    }
    for (long long i = n - 1 - m; i < n - 1; ++i) {
      ans = (ans + (contr[i] * p[i - (n - 1 - m)]) % 1000000007) % 1000000007;
    }
  } else {
    for (long long i = 0; i < n - 2; ++i) {
      ans = (ans + (contr[i] * p[i]) % 1000000007) % 1000000007;
    }
    long long x = 1;
    long long i = n - 2;
    while (i < m) {
      x = (x * p[i]) % 1000000007;
      i++;
    }
    ans = (ans + (contr[n - 2] * x) % 1000000007) % 1000000007;
  }
  cout << ans << endl;
}
int32_t main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```