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 = 1e5 + 5;
const long long mod = 1e9 + 7;
long long t, n, m, p, ans, num;
long long sz[maxn], head[maxn], cnt, v[maxn];
priority_queue<long long> q;
struct edge {
  long long to, next;
} e[maxn << 1];
bool cmp(long long x, long long y) { return x > y; }
void adde(long long u, long long v) {
  e[++cnt].to = v;
  e[cnt].next = head[u];
  head[u] = cnt;
}
void dfs(long long u, long long fa) {
  sz[u] = 1;
  for (register long long i = head[u]; i; i = e[i].next) {
    long long w = e[i].to;
    if (w == fa) continue;
    dfs(w, u);
    sz[u] += sz[w];
    v[++num] = sz[w] * (n - sz[w]);
  }
}
signed main() {
  scanf("%lld", &t);
  while (t--) {
    memset(head, 0, sizeof(head));
    ans = 0;
    cnt = 0;
    num = 0;
    memset(sz, 0, sizeof(sz));
    memset(v, 0, sizeof(v));
    scanf("%lld", &n);
    for (register long long i = 1, u, w; i < n; ++i)
      scanf("%lld%lld", &u, &w), adde(u, w), adde(w, u);
    dfs(1, 0);
    scanf("%lld", &m);
    for (register long long i = 1; i <= m; ++i) scanf("%lld", &p), q.push(p);
    sort(v + 1, v + num + 1, cmp);
    p = q.top();
    q.pop();
    m--;
    while (n - 1 <= m) p = q.top() % mod * p % mod, q.pop(), m--;
    m++;
    while (n - 1 > m) q.push(1ll), m++;
    ans = (ans + v[1] % mod * p % mod) % mod;
    for (register long long i = 2; i < n; ++i) {
      p = q.top(), q.pop();
      ans = (ans + v[i] % mod * p % mod) % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}
