#include <bits/stdc++.h>
using namespace std;
const long long N = 1e5 + 5, K = 20, mod = 1000000007, MOD = 1000000007;
vector<long long> adj[N], ch(N);
priority_queue<long long> pq;
long long n;
void init(long long n) {
  for (long long i = 0; i <= n; i++) {
    adj[i].clear();
    ch[i] = 0;
  }
  while (!pq.empty()) {
    pq.pop();
  }
}
void dfs1(long long v, long long par) {
  ch[v] = 1;
  for (long long u : adj[v]) {
    if (u == par) continue;
    dfs1(u, v);
    ch[v] += ch[u];
  }
}
void dfs2(long long v, long long par) {
  for (long long u : adj[v]) {
    if (u == par) {
      continue;
    } else {
      long long val = (ch[u] * (n - ch[u]));
      pq.push(val);
      dfs2(u, v);
    }
  }
}
void solve() {
  cin >> n;
  init(n);
  long long x, y;
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  vector<long long> fac;
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> x;
    fac.push_back(x);
  }
  while (fac.size() < n - 1) fac.push_back(1);
  sort((fac).begin(), (fac).end());
  while (fac.size() > n - 1) {
    x = fac.back();
    fac.pop_back();
    fac[fac.size() - 1] *= x;
    fac[fac.size() - 1] %= mod;
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (long long i = n - 2; i >= 0; i--) {
    ans = (ans + (fac[i] * pq.top()) % mod) % mod;
    pq.pop();
  }
  cout << ans << endl;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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 = 1e5 + 5, K = 20, mod = 1000000007, MOD = 1000000007;
vector<long long> adj[N], ch(N);
priority_queue<long long> pq;
long long n;
void init(long long n) {
  for (long long i = 0; i <= n; i++) {
    adj[i].clear();
    ch[i] = 0;
  }
  while (!pq.empty()) {
    pq.pop();
  }
}
void dfs1(long long v, long long par) {
  ch[v] = 1;
  for (long long u : adj[v]) {
    if (u == par) continue;
    dfs1(u, v);
    ch[v] += ch[u];
  }
}
void dfs2(long long v, long long par) {
  for (long long u : adj[v]) {
    if (u == par) {
      continue;
    } else {
      long long val = (ch[u] * (n - ch[u]));
      pq.push(val);
      dfs2(u, v);
    }
  }
}
void solve() {
  cin >> n;
  init(n);
  long long x, y;
  for (long long i = 1; i < n; i++) {
    cin >> x >> y;
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  vector<long long> fac;
  long long m;
  cin >> m;
  for (long long i = 0; i < m; i++) {
    cin >> x;
    fac.push_back(x);
  }
  while (fac.size() < n - 1) fac.push_back(1);
  sort((fac).begin(), (fac).end());
  while (fac.size() > n - 1) {
    x = fac.back();
    fac.pop_back();
    fac[fac.size() - 1] *= x;
    fac[fac.size() - 1] %= mod;
  }
  dfs1(1, 0);
  dfs2(1, 0);
  long long ans = 0;
  for (long long i = n - 2; i >= 0; i--) {
    ans = (ans + (fac[i] * pq.top()) % mod) % mod;
    pq.pop();
  }
  cout << ans << endl;
}
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long TESTS = 1;
  cin >> TESTS;
  while (TESTS--) {
    solve();
  }
  return 0;
}

```