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 + 10;
const long long mode = 1e9 + 7;
long long t, n, m, p[maxn], cur, ctt, head[maxn], ct[maxn];
typedef struct a {
  long long to, next;
} per;
per si[maxn * 4];
void add(long long u, long long v) {
  si[++ctt].next = head[u];
  si[ctt].to = v;
  head[u] = ctt;
}
long long siz[maxn];
void dfs(long long now, long long fa) {
  siz[now] = 1;
  for (long long i = head[now]; i != -1; i = si[i].next) {
    long long to = si[i].to;
    if (to != fa) {
      dfs(to, now);
      siz[now] += siz[to];
    }
  }
  if (siz[now] == 1) return;
  for (long long i = head[now]; i != -1; i = si[i].next) {
    long long to = si[i].to;
    if (to != fa) {
      cur++;
      ct[cur] = siz[to] * (n - siz[to]);
    }
  }
}
bool cmp(long long a, long long b) { return a > b; }
signed main() {
  cin >> t;
  while (t--) {
    cin >> n;
    long long u, v;
    ctt = 0;
    cur = 0;
    for (long long i = 0; i <= n * 3 && i < maxn; ++i) {
      head[i] = -1;
      siz[i] = 0;
      ct[i] = 0;
    }
    for (long long i = 1; i < n; ++i) {
      scanf("%lld%lld", &u, &v);
      add(u, v);
      add(v, u);
    }
    cin >> m;
    for (long long i = 1; i <= m; ++i) scanf("%lld", &p[i]);
    dfs(1, 0);
    sort(p + 1, p + 1 + m, cmp);
    sort(ct + 1, ct + 1 + cur, cmp);
    long long ans = 0;
    if (m <= cur) {
      for (long long i = 1; i <= m; ++i) {
        ans += ct[i] * p[i] % mode;
        ans %= mode;
      }
      for (long long i = m + 1; i <= cur; ++i) {
        ans += ct[i];
        ans %= mode;
      }
    } else {
      long long temp = 1;
      for (long long i = 1; i <= m - cur + 1; ++i) {
        temp *= p[i];
        temp %= mode;
      }
      ans += temp * ct[1] % mode;
      ans %= mode;
      for (long long i = 2; i <= cur; ++i) {
        ans += ct[i] * p[m - cur + i] % mode;
        ans %= mode;
      }
    }
    cout << ans << "\n";
  }
  return 0;
}
