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>
#pragma GCC optimize("O3")
using namespace std;
const int N = 100010;
const int MX = 123456;
const int mod = (int)1e9 + 7;
const int base = 97;
const unsigned long long BS1 = 10000019ULL;
const int INF = (1 << 29);
template <class T>
inline void fastScan(T &x) {
  register char c = getchar();
  int neg = 0;
  x = 0;
  for (; (c < 48 || c > 57) && (c != '-'); c = getchar())
    ;
  if (c == '-') {
    neg = 1;
    c = getchar();
  }
  for (; c > 47 && c < 58; c = getchar()) {
    x = (x << 1) + (x << 3) + c - 48;
  }
  if (neg) x = -x;
}
int n, m;
int inp[N];
int dp[N];
vector<int> g[N], vec;
vector<long long> edges;
void dfs(int u, int par) {
  dp[u] = 1;
  for (int i = 0; i < g[u].size(); i++) {
    int v = g[u][i];
    if (v != par) {
      dfs(v, u);
      dp[u] += dp[v];
      int x = dp[v], y = n - dp[v];
      edges.push_back(1ll * x * y);
    }
  }
}
int Solve() {
  dfs(1, -1);
  for (int i = 1; i <= m; i++) {
    vec.push_back(inp[i]);
  }
  while (vec.size() < (n - 1)) {
    vec.push_back(1);
  }
  sort(vec.rbegin(), vec.rend());
  sort(edges.rbegin(), edges.rend());
  int ans = 0;
  long long c, d;
  while (edges.size() > 1) {
    c = edges.back();
    d = vec.back();
    edges.pop_back();
    vec.pop_back();
    ans = (ans + 1ll * c * d % mod) % mod;
  }
  c = edges[0];
  d = 1;
  while (vec.size() > 0) {
    d = 1ll * d * vec.back() % mod;
    vec.pop_back();
  }
  ans = ans + c * d % mod;
  ans %= mod;
  return ans;
}
int main() {
  int test;
  fastScan(test);
  while (test--) {
    fastScan(n);
    for (int i = 1; i <= n; i++) {
      dp[i] = 0;
      g[i].clear();
    }
    vec.clear();
    edges.clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      fastScan(u);
      fastScan(v);
      g[u].push_back(v);
      g[v].push_back(u);
    }
    fastScan(m);
    for (int i = 1; i <= m; i++) {
      fastScan(inp[i]);
    }
    printf("%d\n", Solve());
  }
  return 0;
}
