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;
  void fastIO() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
  }
  vector<int> edge[110];
  int dist[110];
  const long long MOD = 1e9 + 7;
  int cnt[110][110];
  long long dp[110][110][110];
  void dfs(int node, int pa = 0) {
    for (int nx : edge[node]) {
      if (nx == pa) continue;
      dist[nx] = dist[node] + 1;
      dfs(nx, node);
    }
  }
  int main() {
    fastIO();
    int T;
    cin >> T;
    while (T--) {
      int N, K;
      cin >> N >> K;
      for (int i = 1; i <= N; i++) edge[i].clear();
      for (int i = 1; i <= N - 1; i++) {
        int a, b;
        cin >> a >> b;
        edge[a].push_back(b);
        edge[b].push_back(a);
      }
      if (K == 2) {
        cout << (long long)N * (N - 1) / 2 << "\n";
        continue;
      }
      long long tot = 0;
      for (int i = 1; i <= N; i++) {
        for (int j = 0; j < (int)edge[i].size(); j++) {
          for (int k = 0; k <= N; k++) {
            cnt[j][k] = 0;
            dist[k] = 0;
          }
          dist[edge[i][j]] = 1;
          dfs(edge[i][j], i);
          for (int k = 1; k <= N; k++) {
            cnt[j][dist[k]]++;
          }
        }
        for (int j = 0; j <= (int)edge[i].size(); j++) {
          for (int k = 0; k <= K; k++) {
            for (int l = 0; l <= N; l++) {
              dp[j][k][l] = 0;
            }
          }
          for (int l = 0; l <= N; l++) {
            dp[j][0][l] = 1;
          }
        }
        for (int j = 1; j <= (int)edge[i].size(); j++) {
          for (int k = 1; k <= K; k++) {
            for (int l = 1; l <= N; l++) {
              dp[j][k][l] =
                  dp[j - 1][k - 1][l] * cnt[j - 1][l] % MOD + dp[j - 1][k][l];
              dp[j][k][l] %= MOD;
            }
          }
        }
        for (int l = 1; l <= N; l++) {
          tot += dp[(int)edge[i].size()][K][l];
          tot %= MOD;
        }
      }
      cout << tot << "\n";
    }
  }
  