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;
  constexpr int64_t mod = 1e9 + 7;
  constexpr int64_t mxn = 101;
  int64_t d[mxn][mxn];
  vector<int64_t> adj[mxn];
  void dfs(int64_t u, int64_t p, int64_t l, int64_t r) {
    d[r][l]++;
    for (auto& v : adj[u]) {
      if (v != p) {
        dfs(v, u, l + 1, r);
      }
    }
  }
  int64_t dp[mxn][mxn];
  int64_t compute(vector<int64_t>& f, int64_t k) {
    int64_t m = f.size();
    if (m < k) return 0;
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    dp[0][1] = f[0];
    for (int64_t i = 1; i < m; ++i) {
      dp[i][0] = 1;
      for (int64_t j = 1; j <= k; ++j) {
        dp[i][j] = (dp[i - 1][j] + dp[i - 1][j - 1] * f[i] % mod) % mod;
      }
    }
    return dp[m - 1][k];
  }
  void hike() {
    int64_t n, k;
    cin >> n >> k;
    for (int64_t i = 0; i + 1 < n; ++i) {
      int64_t u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << "\n";
      for (int64_t i = 1; i <= n; ++i) {
        adj[i].clear();
      }
      return;
    }
    int64_t count = 0;
    for (int64_t i = 1; i <= n; ++i) {
      for (auto& v : adj[i]) {
        dfs(v, i, 1, v);
      }
      for (int64_t l = 1; l < n; ++l) {
        vector<int64_t> f;
        for (int64_t j = 1; j <= n; ++j) {
          if (d[j][l]) {
            f.push_back(d[j][l]);
          }
        }
        (count += compute(f, k)) %= mod;
      }
      for (int64_t i = 0; i <= n; ++i) {
        for (int64_t j = 0; j <= n; ++j) {
          d[i][j] = 0;
        }
      }
    }
    cout << count << "\n";
    for (int64_t i = 1; i <= n; ++i) {
      adj[i].clear();
    }
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int64_t t = 1;
    cin >> t;
    while (t--) hike();
    return 0;
  }
  