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;
  const int mod = (int)1e9 + 7;
  int ans;
  vector<vector<int> > g;
  vector<int> cnt;
  int mul(int a, int b) { return ((a * 1LL * b) % mod); }
  void add(int& a, int b) {
    a += b;
    if (a >= mod) a -= mod;
  }
  void calculate(int t, int k) {
    vector<vector<int> > dp(t + 1, vector<int>(k + 1));
    dp[0][0] = 1;
    for (int tree_no = 1; tree_no <= t; tree_no++) {
      dp[tree_no][0] = 1;
      for (int sel = 1; sel <= k; sel++) {
        add(dp[tree_no][sel], dp[tree_no - 1][sel]);
        add(dp[tree_no][sel], mul(dp[tree_no - 1][sel - 1], cnt[tree_no]));
      }
    }
    add(ans, dp[t][k]);
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      g.clear();
      g.resize(n);
      ans = 0;
      for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        g[--u].push_back(--v);
        g[v].push_back(u);
      }
      if (k == 2) {
        if (n & 1)
          cout << mul(n, (n - 1) / 2) << '\n';
        else
          cout << mul(n / 2, n - 1) << '\n';
        continue;
      }
      for (int i = 0; i < n; i++) {
        int t = g[i].size();
        vector<int> visited(n);
        cnt.resize(t + 1);
        vector<pair<int, int> > layer;
        visited[i] = 1;
        for (int j = 0; j < t; j++) {
          layer.push_back(make_pair(g[i][j], j + 1));
          cnt[j + 1] = 1;
          visited[g[i][j]] = 1;
        }
        while (!layer.empty()) {
          calculate(t, k);
          vector<pair<int, int> > new_layer;
          while (!layer.empty()) {
            int v = (layer.back()).first;
            int tree_no = (layer.back()).second;
            layer.pop_back();
            cnt[tree_no]--;
            for (int child : g[v]) {
              if (!visited[child]) {
                visited[child] = 1;
                new_layer.push_back(make_pair(child, tree_no));
                cnt[tree_no]++;
              }
            }
          }
          layer = new_layer;
        }
      }
      cout << ans << '\n';
    }
  }
  