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 i64 = int64_t;
  using _ii = pair<int, int>;
  const int N = 107;
  int t, n, k;
  i64 dp_in[N][N], dp_out[N][N], ans, mod = 1e9 + 7;
  vector<vector<int>> a;
  void DFS(int h, int legacy) {
    for (auto &i : a[h]) {
      if (i == legacy) continue;
      DFS(i, h);
      for (int j = 1; j <= n; j++)
        dp_in[h][j] = (dp_in[h][j] + dp_in[i][j - 1]) % mod;
    }
    ++dp_in[h][0];
  }
  void Cal(vector<int> &a, int cnt) {
    int m = a.size();
    vector<i64> f(cnt + 1);
    f[0] = 1;
    for (auto &i : a) {
      for (int j = cnt; j >= 1; j--) {
        f[j] += f[j - 1] * i;
        f[j] %= mod;
      }
    }
    ans = (ans + f[cnt]) % mod;
  }
  void Run(int h, int legacy) {
    for (int j = 1; j <= n; j++) {
      vector<int> d;
      d.push_back(dp_out[h][j]);
      for (auto &i : a[h]) {
        if (i == legacy) continue;
        d.push_back(dp_in[i][j - 1]);
      }
      Cal(d, k);
    }
    for (auto &i : a[h]) {
      if (i == legacy) continue;
      for (int j = 1; j <= n; j++) {
        int temp = dp_in[h][j - 1] - (j - 2 >= 0 ? dp_in[i][j - 2] : 0);
        dp_out[i][j] = temp + dp_out[h][j - 1];
      }
      Run(i, h);
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      cin >> n >> k;
      a.resize(n + 1);
      for (auto &i : a) i.clear();
      for (int i = 1; i <= n - 1; i++) {
        int x, y;
        cin >> x >> y;
        a[x].push_back(y);
        a[y].push_back(x);
      }
      ans = 0;
      memset(dp_in, 0, sizeof(dp_in));
      memset(dp_out, 0, sizeof(dp_out));
      DFS(1, 0);
      Run(1, 0);
      if (k == 2) ans = (n * (n - 1) / 2) % mod;
      cout << ans << '\n';
    }
  }
  