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 mod = 1e9 + 7;
  const int N = 105;
  long long f[N][N];
  long long cnt[N][N];
  vector<int> q[N];
  int maxn = 0;
  void dfs(int rt, int u, int fa, int dep) {
    cnt[rt][dep]++;
    maxn = max(maxn, dep);
    for (auto &v : q[u]) {
      if (v == fa) continue;
      dfs(rt, v, u, dep + 1);
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 0; i <= n; i++) {
        q[i].clear();
      }
      for (int i = 1; i <= n - 1; i++) {
        int u, v;
        cin >> u >> v;
        q[u].push_back(v);
        q[v].push_back(u);
      }
      if (k == 2) {
        cout << (long long)n * (n - 1) % mod / 2 % mod << '\n';
        continue;
      }
      long long ans = 0;
      for (int i = 1; i <= n; i++) {
        memset(cnt, 0, sizeof cnt);
        int mx_d = 0;
        for (auto &u : q[i]) {
          maxn = 0;
          dfs(u, u, i, 0);
          mx_d = max(mx_d, maxn);
        }
        for (int w = 0; w <= mx_d; w++) {
          memset(f, 0, sizeof f);
          f[0][0] = 1;
          f[0][1] = cnt[q[i][0]][w];
          for (int ii = 1; ii < q[i].size(); ii++) {
            f[ii][0] = 1;
            for (int j = 1; j <= min(k, ii + 1); j++)
              f[ii][j] = (f[ii - 1][j] +
                          (long long)f[ii - 1][j - 1] * cnt[q[i][ii]][w] % mod) %
                         mod;
          }
          ans = (ans + f[q[i].size() - 1][k]) % mod;
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  