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>
  const long long N = 205, P = 1000000007;
  long long T, n, k, x, y, num[N][N], f[N][N];
  std::vector<long long> g[N];
  std::vector<long long> a;
  void dfs(long long u, long long fa) {
    num[u][0] = 1;
    for (long long i = 0; i < (long long)g[u].size(); i++) {
      long long v = g[u][i];
      if (v == fa) continue;
      dfs(v, u);
      for (long long j = 1; j <= n; j++) num[u][j] += num[v][j - 1];
    }
  }
  signed main() {
    std::cin >> T;
    while (T--) {
      std::cin >> n >> k;
      long long ans = 0;
      for (long long i = 1; i < n; i++)
        std::cin >> x >> y, g[x].push_back(y), g[y].push_back(x);
      if (k == 2) {
        std::cout << n * (n - 1) / 2 << '\n';
        for (long long i = 1; i <= n; i++) g[i].clear();
        continue;
      }
      for (long long i = 1; i <= n; i++) {
        dfs(i, 0);
        for (long long j = 1; j <= n; j++) {
          for (long long l = 0; l < (long long)g[i].size(); l++)
            a.push_back(num[g[i][l]][j - 1]);
          f[0][0] = 1;
          for (long long x = 1; x <= (long long)a.size(); x++) {
            f[x][0] = 1;
            for (long long y = 1; y <= x; y++)
              f[x][y] = (f[x - 1][y] + f[x - 1][y - 1] * a[x - 1] % P) % P;
          }
          ans = (ans + f[a.size()][k]) % P;
          for (long long x = 1; x <= (long long)a.size(); x++)
            for (long long y = 1; y <= x; y++) f[x][y] = 0;
          a.clear();
        }
        for (long long j = 1; j <= n; j++)
          for (long long l = 0; l <= n; l++) num[j][l] = 0;
      }
      std::cout << ans << '\n';
      for (long long i = 1; i <= n; i++) g[i].clear();
      for (long long i = 1; i <= n; i++)
        for (long long j = 0; j <= n; j++) num[i][j] = 0;
    }
    return 0;
  }
  