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;
  using ll = long long;
  const int MOD = 1e9 + 7;
  int cnt[110];
  vector<int> v[110];
  int dp[110][110];
  ll add(ll x, ll y) { return (x + y) % MOD; }
  ll mul(ll x, ll y) { return x * y % MOD; }
  ll get_ans(int n, int k) {
    for (int i = 0; i <= n; i++) {
      for (int j = 0; j <= k; j++) dp[i][j] = 0;
    }
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
      dp[i][0] = 1;
      for (int j = 1; j <= k; j++) {
        dp[i][j] = add(dp[i - 1][j], dp[i][j]);
        dp[i][j] = add(dp[i][j], mul(dp[i - 1][j - 1], cnt[i]));
      }
    }
    return dp[n][k];
  }
  void dfs(int u, int fa = -1) {
    cout << u << endl;
    for (auto t : v[u]) {
      if (t == fa) continue;
      dfs(t, u);
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; i++) v[i].clear();
      for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        v[a].push_back(b);
        v[b].push_back(a);
      }
      if (k == 2) {
        printf("%lld\n", 1ll * n * (n - 1) / 2 % MOD);
        continue;
      }
      ll ans = 0;
      for (int i = 1; i <= n; i++) {
        vector<int> used(n + 1, 0);
        used[i] = true;
        int m = v[i].size();
        vector<pair<int, int>> layer;
        for (int j = 0; j < m; j++) {
          int y = v[i][j];
          layer.push_back({y, j + 1});
          used[y] = true;
          cnt[j + 1] = 1;
        }
        while (!layer.empty()) {
          ans = add(ans, get_ans(m, k));
          vector<pair<int, int>> newlayer;
          for (auto now : layer) {
            cnt[now.second]--;
            for (auto next : v[now.first]) {
              if (used[next]) continue;
              newlayer.push_back({next, now.second});
              used[next] = true;
              cnt[now.second]++;
            }
          }
          layer = newlayer;
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  