#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 7;
vector<int> g[N];
int n, m, p[N], fa[N];
const long long mod = 1e9 + 7;
int cnt;
vector<long long> v;
long long sz[N];
void dfs(int u, int f) {
  sz[u] = 1;
  fa[u] = f;
  for (int it : g[u]) {
    if (it == f) continue;
    dfs(it, u);
    sz[u] += sz[it];
  }
}
int main() {
  int t;
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
    v.clear();
    cnt = n;
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p[i];
    }
    dfs(1, 0);
    for (int i = n; i; i--) {
      v.push_back(1ll * sz[i] * (n - sz[i]));
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    sort(p + 1, p + m + 1);
    reverse(p + 1, p + m + 1);
    if (m > n - 1) {
      int cnt = m - (n - 1);
      long long base = 1;
      for (int i = 1; i <= cnt + 1; i++) {
        base = base * p[i];
        base %= mod;
      }
      long long ans = (base * v[0] % mod) % mod;
      cnt += 2;
      for (int i = 1; i < v.size(); i++) {
        ans = (ans + (v[i] % mod * p[cnt++]) % mod) % mod;
      }
      cout << ans << endl;
    } else {
      long long ans = 0;
      for (int i = 1; i <= m; i++) {
        ans = (ans + (v[i - 1] % mod * p[i] % mod) % mod) % mod;
      }
      for (int i = m; i < v.size(); i++) {
        ans = (ans + v[i]) % mod;
      }
      cout << ans << endl;
    }
  }
}

### 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;
const int N = 2e5 + 7;
vector<int> g[N];
int n, m, p[N], fa[N];
const long long mod = 1e9 + 7;
int cnt;
vector<long long> v;
long long sz[N];
void dfs(int u, int f) {
  sz[u] = 1;
  fa[u] = f;
  for (int it : g[u]) {
    if (it == f) continue;
    dfs(it, u);
    sz[u] += sz[it];
  }
}
int main() {
  int t;
  ios::sync_with_stdio(false);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) {
      g[i].clear();
    }
    v.clear();
    cnt = n;
    for (int i = 1; i < n; i++) {
      int x, y;
      cin >> x >> y;
      g[x].push_back(y);
      g[y].push_back(x);
    }
    cin >> m;
    for (int i = 1; i <= m; i++) {
      cin >> p[i];
    }
    dfs(1, 0);
    for (int i = n; i; i--) {
      v.push_back(1ll * sz[i] * (n - sz[i]));
    }
    sort(v.begin(), v.end());
    reverse(v.begin(), v.end());
    sort(p + 1, p + m + 1);
    reverse(p + 1, p + m + 1);
    if (m > n - 1) {
      int cnt = m - (n - 1);
      long long base = 1;
      for (int i = 1; i <= cnt + 1; i++) {
        base = base * p[i];
        base %= mod;
      }
      long long ans = (base * v[0] % mod) % mod;
      cnt += 2;
      for (int i = 1; i < v.size(); i++) {
        ans = (ans + (v[i] % mod * p[cnt++]) % mod) % mod;
      }
      cout << ans << endl;
    } else {
      long long ans = 0;
      for (int i = 1; i <= m; i++) {
        ans = (ans + (v[i - 1] % mod * p[i] % mod) % mod) % mod;
      }
      for (int i = m; i < v.size(); i++) {
        ans = (ans + v[i]) % mod;
      }
      cout << ans << endl;
    }
  }
}

```