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;
  using cd = complex<double>;
  const double PI = acos(-1);
  const int N = 109;
  int c[N][N];
  int mod = 1e9 + 7;
  vector<vector<int>> adj;
  int d[N][N];
  void calc(int node, int par, int dep, int root) {
    d[root][dep]++;
    for (int ch : adj[node])
      if (ch != par) calc(ch, node, dep + 1, root);
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    adj = vector<vector<int>>(n);
    for (int i = 0; i < n - 1; i++) {
      int u, v;
      cin >> u >> v;
      adj[u - 1].push_back(v - 1);
      adj[v - 1].push_back(u - 1);
    }
    if (k == 2) {
      cout << c[n][k] << "\n";
      return;
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      memset(d, 0, sizeof d);
      int cnt = adj[i].size();
      vector<int> v;
      for (int ch : adj[i]) {
        calc(ch, i, 0, ch);
        v.push_back(ch);
      }
      if (cnt < k) continue;
      for (int di = 0; di <= n; di++) {
        int dp[cnt + 1][k + 1];
        memset(dp, 0, sizeof dp);
        dp[0][0] = 1;
        for (int i = 1; i <= cnt; i++) {
          dp[i][0] = 1;
          for (int r = 1; r <= k; r++) {
            dp[i][r] =
                (dp[i - 1][r] + 1ll * dp[i - 1][r - 1] * d[v[i - 1]][di] % mod) %
                mod;
          }
        }
        ans += dp[cnt][k];
        ans %= mod;
      }
    }
    cout << ans << "\n";
  }
  int main() {
    cin.tie(0);
    cin.sync_with_stdio(0);
    c[0][0] = 1;
    for (int i = 1; i < N; i++) {
      c[i][0] = 1;
      for (int j = 1; j <= i; j++)
        c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
    }
    int t;
    cin >> t;
    while (t--) solve();
  }
  