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 = 2e5 + 50;
const long long Mod = 1e9 + 7;
struct node {
  long long to, next, w;
} edge[maxn * 2];
struct Edge {
  long long u, v;
} t[maxn];
long long k = 0, head[maxn];
void add(long long u, long long v) {
  edge[++k].to = v;
  edge[k].next = head[u];
  head[u] = k;
}
long long T, p[maxn], m, n;
long long ans = 0, sz[maxn], cnt[maxn];
void dfs(long long x, long long f, long long opt) {
  sz[x] = 1;
  for (long long i = head[x]; i; i = edge[i].next) {
    if (edge[i].to == f) continue;
    dfs(edge[i].to, x, i);
    sz[x] += sz[edge[i].to];
  }
  if (opt == 0) return;
  cnt[opt] = sz[x] * (n - sz[x]);
  if (cnt[opt] == 0 && opt != 0) cnt[opt] = n - 1;
}
int main() {
  cin >> T;
  while (T--) {
    k = 0;
    memset(head, 0, sizeof(head));
    memset(cnt, 0, sizeof(cnt));
    cin >> n;
    for (long long i = 1; i <= n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      add(u, v);
      add(v, u);
    }
    cin >> m;
    for (long long i = 1; i <= m; i++) cin >> p[i];
    sort(p + 1, p + 1 + m);
    long long cnt1 = n - 1 - m;
    if (cnt1 <= 0) cnt1 = 0;
    dfs(1, 1, 0);
    sort(cnt + 1, cnt + 1 + k);
    while (m > n - 1) {
      p[m - 1] = (p[m - 1] * p[m]) % Mod;
      m--;
    }
    long long ans = 0, opt = 0;
    for (long long i = k / 2 + 1; i <= k; i++) {
      if (cnt1) {
        ans = (ans + cnt[i] * 1) % Mod;
        cnt1--;
      } else {
        ans = (ans + cnt[i] * p[++opt] % Mod) % Mod;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
