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;
  int i, j, k, n, m, t, sum, vis[150], x, y, it;
  long long f[105], res, f2[105][105];
  vector<int> v[205];
  void dfs(int x, int fa, int d) {
    vis[x] = 1;
    if (d == j) {
      f[x]++;
      return;
    }
    for (auto i : v[x]) {
      if (i == fa) {
        continue;
      }
      dfs(i, x, d + 1);
      f[x] += f[i];
    }
  }
  int main() {
    scanf("%d", &t);
    while (t--) {
      scanf("%d%d", &n, &m);
      for (i = 1; i <= n; i++) {
        v[i].clear();
      }
      for (i = 1; i < n; i++) {
        scanf("%d%d", &x, &y);
        v[x].push_back(y);
        v[y].push_back(x);
      }
      if (m == 2) {
        printf("%d\n", n * (n - 1) / 2);
        continue;
      }
      res = 0;
      for (i = 1; i <= n; i++) {
        for (j = 1; j <= n; j++) {
          memset(vis, 0, sizeof(vis));
          memset(f, 0, sizeof(f));
          memset(f2, 0, sizeof(f2));
          dfs(i, 0, 0);
          f2[0][0] = 1;
          it = 0;
          for (auto jj : v[i]) {
            if (f[jj] == 0) {
              continue;
            }
            for (k = 0; k <= it; k++) {
              f2[it][k] %= 1000000007;
              f2[it + 1][k] += f2[it][k];
              f2[it + 1][k + 1] += f2[it][k] * f[jj] % 1000000007;
            }
            it++;
          }
          res += f2[it][m];
          res %= 1000000007;
        }
      }
      printf("%lld\n", res);
    }
  }
  