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 int maxn = 1e5 + 100;
const int mod = 1e9 + 7;
int t;
int n;
int m;
long long p[maxn];
int size[maxn];
vector<int> g[maxn];
void dfs(int u, int pre) {
  size[u] = 1;
  for (int v : g[u]) {
    if (v == pre) continue;
    dfs(v, u);
    size[u] += size[v];
  }
}
struct node {
  long long x, y;
  bool operator<(const node &r) const {
    return (long long)x * y > (long long)r.x * r.y;
  }
} Node[maxn];
bool cmp(int x, int y) { return x > y; }
int main() {
  scanf("%d", &t);
  while (t--) {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      g[x].push_back(y);
      g[y].push_back(x);
    }
    scanf("%d", &m);
    for (int i = 1; i <= m; i++) scanf("%lld", p + i);
    sort(p + 1, p + m + 1, cmp);
    dfs(1, 0);
    int tot = 0;
    for (int i = 2; i <= n; i++)
      Node[++tot] = {(long long)n - size[i], (long long)size[i]};
    sort(Node + 1, Node + tot + 1);
    long long ans = 0;
    if (n - 1 >= m) {
      for (int i = 1; i <= m; i++) {
        ans += (long long)p[i] * Node[i].x * Node[i].y;
        ans %= mod;
      }
      for (int i = m + 1; i <= tot; i++) {
        ans += (long long)Node[i].x * Node[i].y;
        ans %= mod;
      }
    } else {
      long long tt = 1;
      for (int i = 1; i <= (m - n + 2); i++) {
        tt *= p[i];
        tt %= mod;
      }
      ans += (long long)tt * Node[1].x % mod * Node[1].y;
      ans %= mod;
      for (int i = m - n + 3; i <= m; i++) {
        ans +=
            (long long)p[i] * Node[i - (m - n + 1)].x * Node[i - (m - n + 1)].y;
        ans %= mod;
      }
    }
    printf("%lld\n", ans);
  }
}
