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;
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<long long> g[n + 1];
    for (long long i = 1; i < n; i++) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 % mod << "\n";
      return;
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      long long len = (long long)g[i].size();
      vector<vector<long long>> cnt(n + 1, vector<long long>(len + 1));
      queue<pair<long long, pair<long long, long long>>> q;
      vector<bool> used(n + 1);
      used[i] = true;
      for (long long j = 0; j < len; j++) {
        used[g[i][j]] = true;
        q.push({g[i][j], {1, j + 1}});
        cnt[1][j + 1]++;
      }
      while (!q.empty()) {
        auto cur = q.front();
        q.pop();
        long long v = cur.first, depth = cur.second.first,
                  subtree = cur.second.second;
        for (long long u : g[v]) {
          if (!used[u]) {
            used[u] = true;
            q.push({u, {depth + 1, subtree}});
            cnt[depth + 1][subtree]++;
          }
        }
      }
      for (long long depth = 1; depth <= n; depth++) {
        vector<vector<long long>> dp(len + 1, vector<long long>(k + 1));
        dp[0][0] = 1;
        for (long long j = 1; j <= len; j++) {
          for (long long vers = 0; vers <= k; vers++) {
            dp[j][vers] += dp[j - 1][vers];
            if (vers > 0) dp[j][vers] += dp[j - 1][vers - 1] * cnt[depth][j];
            dp[j][vers] %= mod;
          }
        }
        ans += dp[len][k];
        ans %= mod;
      }
    }
    cout << ans << "\n";
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  