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;
  using ll = long long;
  using VL = vector<ll>;
  using VVL = vector<VL>;
  using VI = vector<int>;
  using VVI = vector<VI>;
  const ll MOD = 1e9 + 7;
  VVI G;
  VI V;
  VVL M;
  ll f(int i, int j) {
    if (j < 0 or j > i) return 0;
    ll& res = M[i][j];
    if (res != -1) return res;
    if (j == 0) return res = 1;
    return res = (f(i - 1, j) + V[i - 1] * f(i - 1, j - 1)) % MOD;
  }
  void dfs(int x, int p, int d, VI& V) {
    if (d >= (int)V.size())
      V.push_back(1);
    else
      ++V[d];
    for (int y : G[x])
      if (y != p) dfs(y, x, d + 1, V);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      G = VVI(n);
      for (int r = 0; r < n - 1; ++r) {
        int x, y;
        cin >> x >> y;
        --x;
        --y;
        G[x].push_back(y);
        G[y].push_back(x);
      }
      if (k == 2)
        cout << n * (n - 1) / 2 << endl;
      else {
        ll res = 0;
        for (int x = 0; x < n; ++x) {
          int m = G[x].size();
          VVI T(m);
          int mx = 0;
          for (int i = 0; i < m; ++i) {
            dfs(G[x][i], x, 0, T[i]);
            mx = max(mx, (int)T[i].size());
          }
          for (int d = 0; d < mx; ++d) {
            V.clear();
            for (int i = 0; i < m; ++i)
              if (d < (int)T[i].size()) V.push_back(T[i][d]);
            int q = V.size();
            if (q >= k) {
              M = VVL(q + 1, VL(k + 1, -1));
              res = (res + f(q, k)) % MOD;
            }
          }
        }
        cout << res << endl;
      }
    }
  }
  