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;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int T;
    cin >> T;
    while (T--) {
      int n, k;
      cin >> n >> k;
      vector<vector<int>> G(n + 1);
      for (int i = 1; i < n; i++) {
        int x, y;
        cin >> x >> y;
        G[x].push_back(y);
        G[y].push_back(x);
      }
      vector<vector<int>> C(n + 1, vector<int>(n + 1));
      const int MOD = (int)1e9 + 7;
      C[0][0] = 1;
      for (int i = 1; i <= n; i++) {
        C[i][0] = C[i][i] = 1;
        for (int j = 1; j < i; j++) {
          C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
        }
      }
      vector<int> d(n + 1);
      vector<int> f(n + 1);
      function<void(int, int, int)> dfs = [&](int cur, int fa,
                                              int depth) -> void {
        d[cur] = depth;
        f[cur] = fa;
        for (auto nx : G[cur]) {
          if (nx == fa) {
            continue;
          }
          dfs(nx, cur, depth + 1);
        }
      };
      int ans = 0;
      for (int root = 1; root <= n; root++) {
        dfs(root, 0, 0);
        if ((int)G[root].size() < k) {
          continue;
        }
        vector<vector<int>> cnt(G[root].size(), vector<int>(n));
        function<void(int, vector<int>&)> dfs2 = [&](int cur,
                                                     vector<int>& count) -> void {
          count[d[cur]] += 1;
          for (auto nx : G[cur]) {
            if (nx == f[cur]) {
              continue;
            }
            dfs2(nx, count);
          }
        };
        for (int i = 0; i < (int)G[root].size(); i++) {
          dfs2(G[root][i], cnt[i]);
        }
        for (int depth = 1; depth < n; depth++) {
          vector<int> a(G[root].size());
          for (int i = 0; i < (int)G[root].size(); i++) {
            a[i] = cnt[i][depth];
          }
          vector<int> dp(k + 1);
          dp[0] = 1;
          for (int i = 0; i < (int)G[root].size(); i++) {
            for (int j = k; j >= 1; j--) {
              dp[j] = (dp[j] + 1LL * dp[j - 1] * a[i]) % MOD;
            }
          }
          ans = (ans + dp[k]) % MOD;
        }
      }
      if (k == 2) {
        ans = n * (n - 1) / 2;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  