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;
  int d[1000005];
  int f[505][505];
  vector<int> g[1000005];
  void dfs(int u, int r, int pr, int val, int dis = 1) {
    if (dis == val) d[r]++;
    for (auto &v : g[u])
      if (v != pr) dfs(v, r, u, val, dis + 1);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t-- > 0) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; ++i) g[i].clear();
      for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
        continue;
      }
      int res = 0;
      for (int i = 1; i <= n; ++i) {
        if ((g[i]).size() < k) continue;
        for (int j = 1; j < n; ++j) {
          for (auto &v : g[i]) d[v] = 0;
          for (auto &v : g[i]) dfs(v, v, i, j);
          for (int first = 0; first <= (g[i]).size(); ++first)
            for (int second = 0; second <= k; ++second) f[first][second] = 0;
          f[0][0] = 1;
          for (int first = 1; first <= (g[i]).size(); ++first) {
            for (int second = 0; second <= k; ++second) {
              f[first][second] = f[first - 1][second];
              if (second > 0)
                f[first][second] =
                    (f[first][second] + (long long)f[first - 1][second - 1] *
                                            d[g[i][first - 1]] % 1000000007) %
                    1000000007;
            }
          }
          res = (res + f[(g[i]).size()][k]) % 1000000007;
        }
      }
      cout << res << endl;
    }
  }
  