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 pii = pair<long long, long long>;
  const long long INF = 1e9;
  const long long MOD = 1e9 + 7;
  const long long N = 105;
  long long d[N];
  void dfs(long long node, long long par, long long depth,
           vector<vector<long long>>& graph) {
    ++d[depth];
    for (auto i : graph[node]) {
      if (i == par) continue;
      dfs(i, node, depth + 1, graph);
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k;
      cin >> n >> k;
      vector<vector<long long>> graph(n + 1);
      for (long long i = 0; i < n - 1; i++) {
        long long u, v;
        cin >> u >> v;
        graph[u].emplace_back(v);
        graph[v].emplace_back(u);
      }
      if (k == 2) {
        long long ans = (n * (n - 1) / 2) % MOD;
        cout << ans << "\n";
        continue;
      }
      long long ans = 0;
      for (long long root = 1; root <= n; root++) {
        vector<vector<long long>> deps(n + 1);
        for (auto c : graph[root]) {
          memset(d, 0, sizeof d);
          dfs(c, root, 1, graph);
          for (long long i = 1; i <= n; i++) {
            if (d[i]) deps[i].emplace_back(d[i]);
          }
        }
        for (auto dep : deps) {
          long long sz = dep.size();
          if (sz < k) continue;
          vector<vector<long long>> dp(sz + 1, vector<long long>(k + 1));
          dp[0][0] = 1;
          for (long long i = 1; i <= sz; i++) {
            for (long long j = 0; j <= k; j++) {
              dp[i][j] = dp[i - 1][j];
              if (j)
                dp[i][j] = (dp[i][j] + (dp[i - 1][j - 1] * dep[i - 1])) % MOD;
            }
          }
          ans = (ans + dp[sz][k]) % MOD;
        }
      }
      cout << ans << "\n";
    }
  }
  