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>
const int maxn = 1e5 + 7;
const int mod = 1e9 + 7;
struct Edge {
  int t, pre;
  Edge() {}
  Edge(int _t, int _pre) : t(_t), pre(_pre) {}
};
Edge G[maxn * 2];
int head[maxn];
int idx;
int n, m;
long long cnt[maxn];
int MM[60000 + 7];
void init() {
  memset(head, -1, sizeof head);
  idx = 0;
}
void addEdge(int s, int t) {
  Edge in = Edge(t, head[s]);
  G[idx] = in;
  head[s] = idx++;
}
int dfs(int s, int now) {
  int tmp = head[now];
  int ans = 1;
  while (~tmp) {
    Edge& e = G[tmp];
    if (e.t == s) {
      tmp = e.pre;
      continue;
    }
    int nodeNumber = dfs(now, e.t);
    cnt[tmp >> 1] = 1LL * nodeNumber * (n - nodeNumber);
    ans += nodeNumber;
    tmp = e.pre;
  }
  return ans;
}
int solve() {
  init();
  scanf("%d", &n);
  for (int i = 0; i < n - 1; ++i) {
    int s, t;
    scanf("%d%d", &s, &t);
    addEdge(s, t);
    addEdge(t, s);
  }
  scanf("%d", &m);
  for (int i = 0; i < m; i++) scanf("%d", MM + i);
  dfs(-1, 1);
  std::sort(cnt, cnt + n - 1);
  std::sort(MM, MM + m);
  long long ans = 0LL;
  if (n - 1 >= m) {
    int i, j;
    for (i = n - 2, j = m - 1; i >= 0 && j >= 0; --i, --j) {
      ans = ans + cnt[i] * MM[j];
      ans = ans % mod;
    }
    for (; i >= 0; i--) {
      ans = ans + cnt[i];
      ans = ans % mod;
    }
  } else {
    int i, j;
    for (i = 0, j = 0; i < n - 2 && j < m - 1; ++i, ++j) {
      ans = ans + cnt[i] * MM[j];
      ans = ans % mod;
    }
    long long tmp = cnt[i] * MM[j];
    ++i, ++j;
    tmp %= mod;
    for (; j < m; j++) {
      tmp = tmp * MM[j];
      tmp %= mod;
    }
    ans = ans + tmp;
    ans %= mod;
  }
  return (int)ans;
}
int main() {
  int t;
  scanf("%d", &t);
  for (int w = 0; w < t; ++w) {
    printf("%d\n", solve());
  }
  return 0;
}
