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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx,avx2,tune=native")
  template <typename T>
  using Prior = std::priority_queue<T>;
  template <typename T>
  using prior = std::priority_queue<T, std::vector<T>, std::greater<T>>;
  using namespace std;
  const long long md = 1e9 + 7;
  inline long long mod(long long x) { return (x % md + md) % md; }
  int n, k;
  vector<vector<long long>> g;
  int tra(int src, int dir, int step) {
    if (!step) return 1;
    int ret = 0;
    for (auto nbr : g[dir]) {
      if (nbr == src) continue;
      ret += tra(dir, nbr, step - 1);
    }
    return ret;
  }
  long long h(const vector<long long>& cef, int k) {
    const int m = cef.size();
    assert(m >= k);
    vector<vector<long long>> dp(m + 1, vector<long long>(k + 1));
    for (int i = 0; i <= m; ++i) dp[i][0] = 1;
    for (int i = 1; i <= m; ++i)
      for (int j = 1; j <= min(i, k); ++j)
        dp[i][j] = mod(dp[i - 1][j] + mod(dp[i - 1][j - 1] * cef[i - 1]));
    return dp[m][k];
  }
  void solve() {
    cin >> n >> k;
    g = vector<vector<long long>>(n);
    for (auto [u, v] : vector<pair<long long, long long>>(n - 1)) {
      cin >> u >> v, --u, --v;
      g[u].push_back(v);
      g[v].push_back(u);
    };
    ;
    if (k == 2) return cout << (n * (n - 1LL) >> 1) << '\n', void();
    long long ans = 0;
    for (int i = 0; i < n; ++i) {
      int m = g[i].size();
      for (int step = 1; step < n; ++step) {
        vector<long long> cnt;
        for (auto dir : g[i])
          if (int tmp = tra(i, dir, step - 1)) cnt.push_back(tmp);
        if (cnt.size() < k) continue;
        ans = mod(ans + h(cnt, k));
      }
    }
    return cout << ans << '\n', void();
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int _ = 1;
    cin >> _;
    vector<long long> __(_);
    iota(begin(__), end(__), 1);
    for (auto i : __) {
      solve();
    }
    return 0;
  }
  