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;
  mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
  template <class T>
  vector<T> make_unique(vector<T> a) {
    sort((a).begin(), (a).end());
    a.erase(unique((a).begin(), (a).end()), a.end());
    return a;
  }
  const int INF = 1e9;
  const int MOD = 1e9 + 7;
  const int N = 111;
  int q[N], dist[N], root[N];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      vector<vector<int>> adj(n);
      for (int _ = 0; _ < (int)(n - 1); _++) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        adj[u].push_back(v);
        adj[v].push_back(u);
      }
      if (k == 2) {
        cout << (long long int)n * (n - 1) / 2 << endl;
        continue;
      }
      long long int ans = 0;
      for (int s = 0; s < (int)(n); s++) {
        fill_n(dist, n, INF);
        int l = 0, r = 0;
        dist[s] = 0;
        q[r++] = s;
        while (l < r) {
          int u = q[l++];
          for (auto v : adj[u]) {
            if (dist[v] > dist[u] + 1) {
              dist[v] = dist[u] + 1;
              root[v] = (u == s ? v : root[u]);
              q[r++] = v;
            }
          }
        }
        vector<map<int, int>> cnt(n);
        for (int u = 0; u < (int)(n); u++) cnt[dist[u]][root[u]]++;
        for (int d = 0; d < (int)(n); d++) {
          vector<int> poly(n + 1);
          poly[0] = 1;
          for (auto x : cnt[d]) {
            for (int i = (int)(n)-1; i >= ((int)0); i--) {
              poly[i + 1] += (long long int)x.second * poly[i] % MOD;
              if (poly[i + 1] >= MOD) poly[i + 1] -= MOD;
            }
          }
          ans += poly[k];
        }
      }
      cout << ans % MOD << endl;
    }
    return 0;
  }
  