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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  long long MOD = 1e9 + 7;
  int t, n, k, u, v;
  vector<vector<int> > adj;
  vector<int> dep, part;
  void dfs(int i, int p, int d) {
    dep[i] = d;
    for (int nb : adj[i])
      if (nb != p) {
        part[nb] = part[i];
        dfs(nb, i, d + 1);
      }
  }
  long long calc(int i) {
    dep.clear();
    dep.resize(n, 0);
    part.clear();
    part.resize(n, i);
    for (int nb : adj[i]) {
      part[nb] = nb;
      dfs(nb, i, 1);
    }
    long long ans = 0;
    for (int d = (1); d < (n); ++d) {
      map<int, long long> freq;
      for (int i = (0); i < (n); ++i)
        if (dep[i] == d) freq[part[i]]++;
      if (freq.size() < k) break;
      vector<long long> vals;
      for (auto p : freq) vals.push_back(p.second);
      vector<vector<long long> > dp(vals.size(), vector<long long>(k, 0));
      dp[0][0] = vals[0];
      for (int i = (1); i < (vals.size()); ++i) {
        dp[i][0] = (vals[i] + dp[i - 1][0]) % MOD;
        for (int j = (1); j < (k); ++j) {
          dp[i][j] = (dp[i - 1][j] + vals[i] * dp[i - 1][j - 1]) % MOD;
        }
      }
      ans = (ans + dp[vals.size() - 1][k - 1]) % MOD;
    }
    return ans;
  }
  void solve() {
    cin >> n >> k;
    adj.clear();
    adj.resize(n);
    for (int i = (0); i < (n - 1); ++i) {
      cin >> u >> v;
      u--;
      v--;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      return;
    }
    long long ans = 0;
    for (int i = (0); i < (n); ++i) ans = (ans + calc(i)) % MOD;
    cout << ans << '\n';
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cin >> t;
    while (t--) {
      solve();
    }
  }
  