description: A tree is an undirected connected graph without cycles.
  
  You are given a tree of n vertices. Find the number of ways to choose exactly k vertices in this tree (i. e. a k-element subset of vertices) so that all pairwise distances between the selected vertices are equal (in other words, there exists an integer c such that for all u, v (u ≠ v, u, v are in selected vertices) d_{u,v}=c, where d_{u,v} is the distance from u to v).
  
  Since the answer may be very large, you need to output it modulo 10^9 + 7.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10) — the number of test cases. Then t test cases follow.
  
  Each test case is preceded by an empty line.
  
  Each test case consists of several lines. The first line of the test case contains two integers n and k (2 ≤ k ≤ n ≤ 100) — the number of vertices in the tree and the number of vertices to be selected, respectively. Then n - 1 lines follow, each of them contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) which describe a pair of vertices connected by an edge. It is guaranteed that the given graph is a tree and has no loops or multiple edges.
  
  Output
  
  For each test case output in a separate line a single integer — the number of ways to select exactly k vertices so that for all pairs of selected vertices the distances between the vertices in the pairs are equal, modulo 10^9 + 7 (in other words, print the remainder when divided by 1000000007).
  
  Example
  
  Input
  
  
  3
  
  4 2
  1 2
  2 3
  2 4
  
  3 3
  1 2
  2 3
  
  5 3
  1 2
  2 3
  2 4
  4 5
  
  
  Output
  
  
  6
  0
  1
solution: #include <bits/stdc++.h>
  using namespace std;
  void _print(long long t) { cerr << t; }
  void _print(string t) { cerr << t; }
  void _print(char t) { cerr << t; }
  void _print(long double t) { cerr << t; }
  void _print(double t) { cerr << t; }
  void _print(unsigned long long t) { cerr << t; }
  template <class T, class V>
  void _print(pair<T, V> p);
  template <class T>
  void _print(vector<T> v);
  template <class T>
  void _print(set<T> v);
  template <class T, class V>
  void _print(map<T, V> v);
  template <class T>
  void _print(multiset<T> v);
  template <class T, class V>
  void _print(pair<T, V> p) {
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
  }
  template <class T>
  void _print(vector<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(set<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T>
  void _print(multiset<T> v) {
    cerr << "[ ";
    for (T i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  template <class T, class V>
  void _print(map<T, V> v) {
    cerr << "[ ";
    for (auto i : v) {
      _print(i);
      cerr << " ";
    }
    cerr << "]";
  }
  const long long N = 102;
  long long n, k;
  vector<long long> adj[N];
  void dfs(long long node, long long par, long long sub_index, long long dep,
           long long &max_dep, vector<vector<long long>> &cnt) {
    cnt[dep][sub_index]++;
    max_dep = max(max_dep, dep);
    for (auto i : adj[node]) {
      if (i == par) continue;
      ;
      dfs(i, node, sub_index, dep + 1, max_dep, cnt);
    }
  }
  long long rec(long long pos, long long setsize, long long &dep, long long &root,
                vector<vector<long long>> &cnt, vector<vector<long long>> &dp) {
    if (pos == (long long)adj[root].size()) {
      return setsize == k;
    }
    if (dp[pos][setsize] != -1) return dp[pos][setsize];
    long long ans = 0;
    if (setsize < k)
      ans = (ans % 1000000007 + (rec(pos + 1, setsize + 1, dep, root, cnt, dp) %
                                 1000000007 * cnt[dep][pos] % 1000000007) %
                                    1000000007) %
            1000000007;
    ans = (ans % 1000000007 +
           rec(pos + 1, setsize, dep, root, cnt, dp) % 1000000007) %
          1000000007;
    return dp[pos][setsize] = ans;
  }
  void solve() {
    cin >> n >> k;
    for (long long i = 0; i <= n; i++) adj[i].clear();
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      long long ans = (n * (n - 1) / 2) % 1000000007;
      cout << ans << "\n";
      return;
    }
    long long ans = 0;
    for (long long root = 0; root < n; root++) {
      vector<vector<long long>> cnt(
          n + 1, vector<long long>((long long)adj[root].size() + 1));
      long long sub_index = 0;
      long long max_dep = 1;
      for (auto i : adj[root]) {
        dfs(i, root, sub_index, 1, max_dep, cnt);
        sub_index++;
      }
      for (long long i = 1; i <= max_dep; i++) {
        vector<vector<long long>> dp((long long)adj[root].size(),
                                     vector<long long>(k + 1, -1));
        long long dep = i;
        ans = (ans % 1000000007 + rec(0, 0, dep, root, cnt, dp) % 1000000007) %
              1000000007;
      }
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t = 1;
    cin >> t;
    while (t--) {
      solve();
    }
  }
  