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 long long mod = 1e9 + 7;
  long long dp[2000];
  int binpow(long long base, long long pow) {
    if (pow == 0) return 1;
    long long half = binpow(base, pow / 2);
    long long ret = (half * half) % mod;
    if (pow % 2 == 1) ret *= base;
    return ret % mod;
  }
  long long choose(long long a, long long b) {
    if (a < 0 || b < 0 || a < b) return 0;
    long long ret = dp[a];
    long long denom = (dp[a - b] * dp[b]) % mod;
    ret *= binpow(denom, mod - 2);
    ret %= mod;
    return ret;
  }
  int main() {
    dp[0] = 1;
    for (int i = 1; i < 2000; i++) {
      dp[i] = (dp[i - 1] * i) % mod;
    }
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int T, N, K, u, v;
    cin >> T;
    while (T--) {
      cin >> N >> K;
      long long ret = 0;
      vector<int> adj[N + 1];
      for (int i = 0; i < N - 1; i++) {
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
      }
      for (int i = 1; i <= N; i++) {
        if (adj[i].size() < K) continue;
        vector<pair<long long, int> > bfs;
        vector<int> vis;
        vis.resize(N + 1);
        vis[i] = 1;
        for (int n : adj[i]) {
          bfs.push_back(pair<long long, int>(n, n));
        }
        while (!bfs.empty()) {
          unordered_map<int, int> sz;
          vector<pair<long long, int> > step;
          for (pair<long long, int> j : bfs) {
            sz[j.second]++;
            vis[j.first] = 1;
            for (int n : adj[j.first]) {
              if (!vis[n]) step.push_back(pair<long long, int>(n, j.second));
            }
          }
          if (sz.size() < K) {
            break;
          }
          int second = sz.size();
          vector<int> all;
          for (auto i : sz) {
            all.push_back(i.second);
          }
          long long dp[second + 1][K + 1];
          for (int i = 0; i <= second; i++) {
            for (int j = 0; j <= K; j++) {
              dp[i][j] = (j == 0);
            }
          }
          for (int i = 1; i <= second; i++) {
            for (int j = 1; j <= K; j++) {
              dp[i][j] = dp[i - 1][j];
              if (j != 0) dp[i][j] += dp[i - 1][j - 1] * all[i - 1];
              dp[i][j] %= mod;
            }
          }
          ret += dp[second][K];
          ret %= mod;
          swap(step, bfs);
        }
      }
      cout << (K == 2 ? (N) * (N - 1) / 2 : ret) << '\n';
    }
    return 0;
  }
  