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 <class T>
inline void read(T &ret) {
  char c;
  int sgn;
  if (c = getchar(), c == EOF) return;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  ret *= sgn;
  return;
}
inline void outi(int x) {
  if (x > 9) outi(x / 10);
  putchar(x % 10 + '0');
}
inline void outl(long long x) {
  if (x > 9) outl(x / 10);
  putchar(x % 10 + '0');
}
const int maxn = 100010;
const int mod = 1e9 + 7;
long long sz[maxn], a[maxn], p[maxn];
vector<long long> edge[maxn];
int cntt = 0, n, m;
void dfs(int u, int pre) {
  sz[u] = 1;
  for (int i = 0; i < edge[u].size(); i++) {
    int v = edge[u][i];
    if (v != pre) {
      dfs(v, u);
      sz[u] += sz[v];
      a[cntt++] = (n - sz[v]) * sz[v];
    }
  }
}
int main() {
  int t;
  read(t);
  while (t--) {
    read(n);
    for (int i = 0; i <= n; i++) edge[i].clear();
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      read(u), read(v);
      edge[u].push_back(v);
      edge[v].push_back(u);
    }
    cntt = 0;
    dfs(1, -1);
    sort(a, a + cntt);
    read(m);
    for (int i = 0; i < m; i++) read(p[i]);
    sort(p, p + m);
    if (cntt > m) {
      for (int i = m; i < cntt; i++) p[i] = 1;
      sort(p, p + cntt);
    } else {
      for (int i = cntt; i < m; i++) p[cntt - 1] = p[cntt - 1] * p[i] % mod;
    }
    long long ans = 0;
    for (int i = cntt - 1; i >= 0; i--) {
      ans = (ans + a[i] % mod * p[i]) % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
