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>
  #pragma GCC optimize("-Ofast", "-funroll-all-loops")
  using namespace std;
  const int MAXN = 1e2 + 10, mod = 1e9 + 7;
  inline int add(int first, int second) {
    return first + second > mod ? first + second - mod : first + second;
  }
  inline int mul(int first, int second) { return 1ll * first * second % mod; }
  int ch[MAXN][MAXN], dp[MAXN][MAXN];
  vector<int> G[MAXN];
  void dfs(int first, int prev, int d, int j) {
    ch[j][d] += 1;
    for (int(i) = (0); (i) < (G[first].size()); (i) += (1)) {
      if (G[first][i] == prev) continue;
      dfs(G[first][i], first, d + 1, j);
    }
  }
  int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n, t, k, ans, a, b;
    bool f;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      ans = 0;
      for (int(i) = (1); (i) <= (n); (i) += (1)) G[i].clear();
      for (int(i) = (1); (i) < (n); (i) += (1)) {
        cin >> a >> b;
        G[a].push_back(b);
        G[b].push_back(a);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        continue;
      }
      for (int(i) = (1); (i) <= (n); (i) += (1)) {
        memset(ch, 0, sizeof(ch));
        if (G[i].size() < k) continue;
        for (int(j) = (0); (j) < (G[i].size()); (j) += (1)) {
          dfs(G[i][j], i, 1, j + 1);
        }
        for (int(z) = (1); (z) < (n); (z) += (1)) {
          memset(dp, 0, sizeof(dp));
          dp[0][0] = 1;
          for (int(j) = (1); (j) <= (G[i].size()); (j) += (1)) {
            for (int(u) = (0); (u) <= (min(j, k)); (u) += (1)) {
              if (u) dp[j][u] = add(dp[j][u], mul(dp[j - 1][u - 1], ch[j][z]));
              dp[j][u] = add(dp[j - 1][u], dp[j][u]);
            }
          }
          ans = add(ans, dp[G[i].size()][k]);
        }
      }
      cout << ans << '\n';
    }
  }
  