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;
template <typename T>
void rd(T& x) {
  long long f = 0, c;
  while (!isdigit(c = getchar())) f ^= !(c ^ 45);
  x = (c & 15);
  while (isdigit(c = getchar())) x = x * 10 + (c & 15);
  if (f) x = -x;
}
template <typename T>
void pt(T x, long long c = -1) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) pt(x / 10);
  putchar(x % 10 + 48);
  if (c != -1) putchar(c);
}
const long long N = 100005, P = 1e9 + 7;
long long T, n, sz[N];
vector<long long> G[N];
void dfs(long long k1, long long k2) {
  sz[k1] = 1;
  for (auto j : G[k1])
    if (j != k2) {
      dfs(j, k1);
      sz[k1] += sz[j];
    }
}
signed main() {
  rd(T);
  while (T--) {
    rd(n);
    for (long long i = (1); i <= (n); ++i) G[i].clear();
    for (long long i = (2); i <= (n); ++i) {
      long long k1, k2;
      rd(k1), rd(k2);
      G[k1].push_back(k2), G[k2].push_back(k1);
    }
    long long m;
    rd(m);
    vector<long long> p(m, 1);
    for (long long i = (0); i <= (m - 1); ++i) {
      rd(p[i]);
    }
    sort(p.begin(), p.end());
    while ((long long)p.size() > n - 1) {
      long long t = p.back();
      p.pop_back();
      (p.back() *= t) %= P;
    }
    while ((long long)p.size() < n - 1) {
      p.push_back(1);
    }
    sort(p.begin(), p.end(),
         [&](long long k1, long long k2) { return k1 > k2; });
    dfs(1, 0);
    vector<long long> num;
    for (long long i = (2); i <= (n); ++i) num.push_back(sz[i] * (n - sz[i]));
    sort(num.begin(), num.end(),
         [&](long long k1, long long k2) { return k1 > k2; });
    long long ans = 0;
    for (long long i = 0; i < (long long)num.size(); ++i)
      (ans += num[i] % P * p[i] % P) %= P;
    pt(ans, '\n');
  }
  return 0;
}
