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;
  int c[101][101];
  int C(int n, int r) {
    if (n == r || r == 0) return 1;
    if (~c[n][r]) return c[n][r];
    return c[n][r] = (C(n - 1, r - 1) + C(n - 1, r)) % mod;
  }
  void solve(int tc) {
    int n, k;
    cin >> n >> k;
    vector<vector<int>> g(n + 1);
    vector<int> deg(n + 1);
    for (int i = 1; i < n; ++i) {
      int u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k <= 2) {
      cout << C(n, k) << '\n';
      return;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      if (g[i].size() >= k) {
        vector<vector<int>> dep(g[i].size(), vector<int>(n + 1));
        vector<int> dep_cnt(n + 1, 0);
        for (int j = 0; j < g[i].size(); ++j) {
          function<void(int, int, int)> dfs = [&](int u, int p, int d) {
            dep[j][d]++;
            for (int v : g[u])
              if (v != p) dfs(v, u, d + 1);
          };
          dfs(g[i][j], i, 1);
        }
        for (int d = 1; d <= n; ++d) {
          vector<vector<int>> dp(k, vector<int>(g[i].size(), 0));
          int cnt = 0;
          for (int j = 0; j < g[i].size(); ++j) {
            cnt += !!dep[j][d];
            dp[0][j] = dep[j][d];
          }
          if (cnt < k) break;
          for (int l = 1; l < k; ++l) {
            long long sum = 0;
            for (int j = l; j < g[i].size(); ++j) {
              sum = (sum + dp[l - 1][j - 1]) % mod;
              dp[l][j] = sum * dp[0][j] % mod;
            }
          }
          for (int j = k - 1; j < g[i].size(); ++j)
            ans = (ans + dp[k - 1][j]) % mod;
        }
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    memset(c, -1, sizeof c);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; ++i) solve(i);
    return 0;
  }
  