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 = 1e2 + 5;
  vector<int> g[N], a[N];
  long long ans, cur[N], f[N][N], mod = 1e9 + 7;
  void dfs(int u, int p, int l) {
    cur[l]++;
    for (int i = 0; i < g[u].size(); i++) {
      int v = g[u][i];
      if (v == p) continue;
      dfs(v, u, l + 1);
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      ans = 0;
      for (int i = 1; i <= n; i++) {
        g[i].clear();
        cur[i] = 0;
        a[i].clear();
      }
      for (int i = 1; i < n; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
        continue;
      }
      for (int r = 1; r <= n; r++) {
        for (int i = 1; i <= n; i++) {
          a[i].clear();
        }
        if (g[r].size() < k) continue;
        for (int i = 0; i < g[r].size(); i++) {
          for (int i = 1; i <= n; i++) {
            cur[i] = 0;
          }
          dfs(g[r][i], r, 1);
          for (int i = 1; i <= n; i++) {
            a[i].push_back(cur[i]);
          }
        }
        for (int i = 1; i <= n; i++) {
          f[0][0] = 1;
          for (int j = 1; j <= a[i].size(); j++) {
            for (int l = 0; l <= min(k, j); l++) {
              f[j][l] = f[j - 1][l] % mod;
              if (l > 0 && a[i][j - 1] > 0)
                f[j][l] = (f[j][l] + f[j - 1][l - 1] * a[i][j - 1]) % mod;
            }
          }
          ans = (ans + f[a[i].size()][k]) % mod;
        }
      }
      cout << ans << endl;
    }
  }
  