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 N = 105;
  const int mod = 1000000007;
  int dp[N][N], cnt[N];
  vector<int> g[N];
  void dfs(int v, int p, int d) {
    cnt[d]++;
    for (int u : g[v]) {
      if (u != p) dfs(u, v, d + 1);
    }
  }
  inline void add(int &x, int y) {
    x += y;
    if (x >= mod) x -= mod;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; i++) {
        g[i].clear();
      }
      for (int i = 2; i <= n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        continue;
      }
      int answer = 0;
      for (int root = 1; root <= n; root++) {
        if (g[root].size() < k) continue;
        memset(dp, 0, sizeof(dp));
        for (int d = 1; d <= n; d++) {
          dp[d][0] = 1;
        }
        for (int v : g[root]) {
          memset(cnt, 0, sizeof(cnt));
          dfs(v, root, 1);
          for (int d = 1; d <= n; d++) {
            for (int cur = k - 1; cur >= 0; cur--) {
              add(dp[d][cur + 1], dp[d][cur] * 1LL * cnt[d] % mod);
            }
          }
        }
        for (int d = 1; d <= n; d++) {
          add(answer, dp[d][k]);
        }
      }
      cout << answer << '\n';
    }
    return 0;
  }
  