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 = 100005;
using ll = long long;
ll mod = 1e9 + 7;
vector<long long> edges[maxn];
long long sub[maxn], depth[maxn];
void dfs(long long x, long long p) {
  sub[x] = 1;
  depth[x] = depth[p] + 1;
  for (auto i : edges[x]) {
    if (i != p) {
      dfs(i, x);
      sub[x] += sub[i];
    }
  }
}
int32_t main(int32_t argc, char const *argv[]) {
  long long t;
  scanf("%lld", &t);
  while (t--) {
    long long n;
    scanf("%lld", &n);
    for (long long i = 1; i <= n; ++i) {
      edges[i].clear();
      sub[i] = depth[i] = 0;
    }
    vector<pair<long long, long long>> inp;
    for (long long i = 0; i < n - 1; ++i) {
      long long x, y;
      scanf("%lld %lld", &x, &y);
      edges[x].push_back(y);
      edges[y].push_back(x);
      inp.push_back({x, y});
    }
    dfs(1, 1);
    vector<pair<long long, long long>> E;
    long long track = 0;
    for (auto i : inp) {
      pair<long long, long long> ret;
      if (depth[i.first] > depth[i.second]) swap(i.first, i.second);
      ret.first = sub[i.second];
      ret.second = n - ret.first;
      if (ret.second > ret.first) swap(ret.first, ret.second);
      E.push_back(ret);
      ++track;
    }
    sort(E.begin(), E.end(),
         [&](pair<long long, long long> x, pair<long long, long long> y) {
           return (ll)x.first * (ll)x.second * 1LL >
                  (ll)y.first * (ll)y.second * 1LL;
         });
    vector<ll> primes;
    long long k;
    scanf("%lld", &k);
    primes.resize(k);
    for (long long i = 0; i < k; ++i) {
      scanf("%lld", &primes[i]);
    }
    sort(primes.begin(), primes.end());
    ll ans = 0;
    long long cnt = 0;
    if (k > n - 1) {
      cnt = k - (n - 1);
    }
    long long cur = k - 1;
    ll res = 0;
    for (long long i = 0; i < n - 1; ++i) {
      if (i == 0) {
        ll ans = (E[i].first * E[i].second) % mod;
        for (long long j = 0; j < cnt; ++j) {
          ans = (ans * primes[cur]) % mod;
          --cur;
        }
        if (cur >= 0) {
          ans = (ans * primes[cur]) % mod;
          --cur;
        }
        res = (res + ans) % mod;
      } else {
        ll ans = (E[i].first * E[i].second) % mod;
        if (cur >= 0) {
          ans = (ans * primes[cur]) % mod;
          --cur;
        }
        res = (res + ans) % mod;
      }
    }
    cout << res << endl;
  }
  return 0;
}
