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 M = 1e9 + 7;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, z, a, b;
      cin >> n >> z;
      vector<int> c(n + 1);
      vector<vector<int>> v(n + 1);
      for (int i = 0; i < n - 1; i++) {
        cin >> a >> b;
        c[a]++, c[b]++;
        v[a].push_back(b);
        v[b].push_back(a);
      }
      if (z == 2)
        cout << (n * (n - 1)) / 2 << '\n';
      else {
        long long int ans = 0;
        for (int i = 1; i <= n; i++) {
          if (c[i] >= z) {
            vector<vector<long long int>> m(n + 1);
            for (int j = 0; j < (int)v[i].size(); j++) {
              queue<pair<int, int>> q;
              vector<int> check(n + 1), cd(n + 1);
              q.push({v[i][j], 1});
              check[v[i][j]] = 1;
              check[i] = 1;
              while (!q.empty()) {
                pair<int, int> atual = q.front();
                q.pop();
                cd[atual.second]++;
                for (int k = 0; k < (int)v[atual.first].size(); k++) {
                  int prox = v[atual.first][k];
                  if (!check[prox]) {
                    check[prox] = 1;
                    q.push({prox, atual.second + 1});
                  }
                }
              }
              for (int k = 1; k <= n; k++) {
                if (cd[k]) {
                  m[k].push_back(cd[k]);
                }
              }
            }
            for (int j = 1; j <= n; j++) {
              if ((int)m[j].size() >= z) {
                vector<vector<int>> dp(((int)m[j].size() + 1),
                                       vector<int>(z + 1));
                for (int k = 0; k <= (int)m[j].size(); k++) dp[k][z] = 1;
                for (int k = z - 1; k >= 0; k--) {
                  for (int l = (int)m[j].size() - 1; l >= 0; l--) {
                    dp[l][k] += (m[j][l] * dp[l + 1][k + 1]) % M;
                    dp[l][k] = (dp[l][k] + dp[l + 1][k]) % M;
                  }
                }
                ans = (ans + dp[0][0]) % M;
              }
            }
          }
        }
        cout << ans << '\n';
      }
    }
    return 0;
  }
  