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 MAX_N = 128;
  const long long mod = 1000 * 1000 * 1000 + 7;
  long long add(long long x, long long y) { return (x + y) % mod; }
  long long mul(long long x, long long y) { return x * y % mod; }
  vector<int> g[MAX_N];
  bool used[MAX_N];
  int cnt[MAX_N];
  long long dp[MAX_N][MAX_N];
  long long rundp(int m, int k) {
    for (int i = 0; i <= m; i++)
      for (int j = 0; j <= k; j++) dp[i][j] = 0;
    dp[0][0] = 1;
    for (int i = 0; i < m; i++)
      for (int j = 0; j <= k; j++) {
        dp[i + 1][j] = add(dp[i + 1][j], dp[i][j]);
        dp[i + 1][j + 1] = add(dp[i + 1][j + 1], mul(dp[i][j], cnt[i]));
      }
    return dp[m][k];
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      g[--a].push_back(--b);
      g[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1LL) / 2 % mod << '\n';
      return;
    }
    long long ans = 0;
    for (int center = 0; center < n; center++) {
      memset(used, 0, n);
      used[center] = true;
      vector<pair<int, int>> layer;
      int m = g[center].size();
      for (int i = 0; i < m; i++) {
        int y = g[center][i];
        layer.emplace_back(y, i);
        cnt[i] = 1;
        used[y] = true;
      }
      while (!layer.empty()) {
        ans = add(ans, rundp(m, k));
        vector<pair<int, int>> newlayer;
        for (auto p : layer) {
          cnt[p.second]--;
          for (auto y : g[p.first])
            if (!used[y]) {
              newlayer.emplace_back(y, p.second);
              used[y] = true;
              cnt[p.second]++;
            }
        }
        layer = newlayer;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  