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.
#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 5;
const long long MAXM = 1e5 + 5;
const long long MOD = 1e9 + 7;
long long siz[MAXN], cnt[MAXN];
long long prm[MAXM];
vector<long long> adj[MAXN];
long long n, m;
void Dfs(long long u, long long fa) {
  siz[u] = 1;
  for (long long v : adj[u]) {
    if (v != fa) {
      Dfs(v, u);
      siz[u] += siz[v];
    }
  }
  cnt[u] = siz[u] * (n - siz[u]);
}
void Solve() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    adj[i].clear();
  }
  for (long long i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  scanf("%lld", &m);
  for (long long i = 1; i <= m; i++) {
    scanf("%lld", prm + i);
  }
  sort(prm + 1, prm + 1 + m);
  if (m >= n) {
    for (long long i = n; i <= m; i++) {
      prm[n - 1] = prm[n - 1] * prm[i] % MOD;
    }
  } else {
    for (long long i = m + 1; i <= n; i++) {
      prm[i] = 1;
    }
    sort(prm + 1, prm + n);
  }
  Dfs(1, 0);
  sort(cnt + 1, cnt + 1 + n);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    cnt[i] %= MOD;
    prm[i] %= MOD;
  }
  for (long long i = 2; i <= n; i++) {
    ans = (ans + cnt[i] * prm[i - 1]) % MOD;
  }
  printf("%lld\n", ans);
}
signed main() {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    Solve();
  }
}
