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;
  long long t;
  long long k, n;
  long long u_, v_;
  vector<long long> g[107];
  long long d[107][107];
  long long f[107][107];
  long long v[107];
  long long dfs(long long root, long long cur, long long dis) {
    d[root][dis]++;
    v[cur] = 1;
    for (auto nxt : g[cur]) {
      if (v[nxt] == 0) {
        dfs(root, nxt, dis + 1);
      }
    }
  }
  signed main() {
    cin >> t;
    while (t--) {
      cin >> n >> k;
      long long ans = 0;
      for (long long i = 1; i <= n; ++i) {
        g[i].clear();
      }
      for (long long i = 1; i <= n - 1; ++i) {
        cin >> u_ >> v_;
        g[u_].push_back(v_);
        g[v_].push_back(u_);
      }
      if (k == 2) {
        cout << ((n * (n - 1)) / 2) << endl;
        continue;
      }
      for (long long i = 1; i <= n; ++i) {
        if (g[i].size() >= k) {
          memset(d, 0, sizeof(d));
          memset(v, 0, sizeof(v));
          v[i] = 1;
          for (long long j = 0; j < g[i].size(); ++j) {
            dfs(j + 1, g[i][j], 1);
          }
          for (long long dep = 1; dep <= n; dep++) {
            for (long long j = 0; j <= g[i].size(); j++) {
              f[j][0] = 1;
            }
            for (long long j = 1; j <= g[i].size(); j++) {
              for (long long k = 1; k <= j; k++) {
                f[j][k] =
                    (f[j - 1][k] + (f[j - 1][k - 1] * d[j][dep]) % 1000000007) %
                    1000000007;
              }
            }
            ans += f[g[i].size()][k];
            ans %= 1000000007;
          }
        }
      }
      cout << ans << endl;
    }
  }
  