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;
void xXx(int x) { cerr << x; }
void xXx(long x) { cerr << x; }
void xXx(long long x) { cerr << x; }
void xXx(unsigned x) { cerr << x; }
void xXx(unsigned long x) { cerr << x; }
void xXx(unsigned long long x) { cerr << x; }
void xXx(float x) { cerr << x; }
void xXx(double x) { cerr << x; }
void xXx(long double x) { cerr << x; }
void xXx(char x) { cerr << '\'' << x << '\''; }
void xXx(const char *x) { cerr << '\"' << x << '\"'; }
void xXx(const string &x) { cerr << '\"' << x << '\"'; }
void xXx(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void xXx(const pair<T, V> &x) {
  cerr << '{';
  xXx(x.first);
  cerr << ',';
  xXx(x.second);
  cerr << '}';
}
template <typename T>
void xXx(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), xXx(i);
  cerr << "}";
}
void xxx() { cerr << "]\n"; }
template <typename T, typename... V>
void xxx(T t, V... v) {
  xXx(t);
  if (sizeof...(v)) cerr << ", ";
  xxx(v...);
}
const int MaxN = 1e5 + 5;
const int Mod = 1e9 + 7;
long long zaed(long long x, long long y) { return (x + y) % Mod; }
long long darb(long long x, long long y) { return (x * y) % Mod; }
int n, m;
vector<int> G[MaxN];
vector<long long> S;
vector<long long> score;
void Init() {
  for (int i = 0; i <= n; ++i) {
    G[i].clear();
  }
  S.resize(n + 1);
  score.clear();
}
void Calc(int x = 1, int p = -1) {
  S[x] = 1;
  for (auto v : G[x]) {
    if (v == p) continue;
    Calc(v, x);
    S[x] += S[v];
    score.push_back(1ll * (n - S[v]) * S[v]);
  }
}
int Go() {
  scanf("%d", &n);
  Init();
  for (int i = 0; i < n - 1; ++i) {
    int x, y;
    scanf("%d %d", &x, &y);
    G[x].push_back(y);
    G[y].push_back(x);
  }
  scanf("%d", &m);
  vector<long long> P(m);
  for (int i = 0; i < m; ++i) scanf("%lld", &P[i]);
  Calc();
  assert(score.size() == n - 1);
  sort(score.begin(), score.end());
  reverse(score.begin(), score.end());
  sort(P.begin(), P.end());
  reverse(P.begin(), P.end());
  long long Ans = 0;
  int len = max(m - (n - 1) + 1, 1);
  int j = 0;
  while (j < len) score[0] = darb(score[0], P[j++]);
  for (int i = 1; i < n - 1 and j < m; ++i, ++j) {
    score[i] = darb(score[i], P[j]);
  }
  for (int i = 0; i < n - 1; ++i) Ans = zaed(Ans, score[i]);
  cout << Ans << endl;
  return 0;
}
int main() {
  int t;
  if (1)
    scanf("%d", &t);
  else
    t = 1;
  while (t--) Go();
}
