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>
  #pragma GCC optimize "O3"
  using namespace std;
  const int INF = 1e9 + 7;
  const int N = 1e6 + 7;
  const int T = 1 << 20;
  vector<vector<int> > G;
  vector<int> przesun(vector<int> a) {
    vector<int> ret((int)a.size() + 1);
    for (int i = 0; i < (int)a.size(); i++) ret[i + 1] = a[i];
    return ret;
  }
  vector<int> operator+(vector<int> a, vector<int> b) {
    int n = max((int)a.size(), (int)b.size());
    a.resize(n);
    b.resize(n);
    vector<int> ret(n);
    for (int i = 0; i < n; i++) ret[i] = a[i] + b[i];
    return ret;
  }
  long long ans = 0;
  int n, k;
  int dp[101][101];
  vector<int> dfs(int start, int prz = -1) {
    vector<vector<int> > kubki;
    for (auto& u : G[start])
      if (u != prz) kubki.push_back(przesun(dfs(u, start)));
    if (prz != -1) {
      vector<int> ret = {1};
      for (auto& u : kubki) ret = ret + u;
      return ret;
    } else {
      for (int odl = 1; odl < n; odl++) {
        for (int i = 0; i <= 100; i++)
          for (int j = 0; j <= 100; j++) dp[i][j] = 0;
        dp[0][0] = 1;
        for (int i = 0; i < (int)kubki.size(); i++) {
          for (int j = 0; j <= i; j++) dp[i + 1][j] = dp[i][j];
          for (int j = 0; j <= i; j++)
            dp[i + 1][j + 1] =
                (dp[i + 1][j + 1] +
                 1ll * dp[i][j] *
                     ((int)kubki[i].size() > odl ? kubki[i][odl] : 0)) %
                INF;
        }
        ans = (ans + dp[(int)kubki.size()][k]) % INF;
      }
    }
    return {};
  }
  int dist[1004][1004];
  void solve() {
    cin >> n >> k;
    ans = 0;
    G = vector<vector<int> >(n + 1);
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      G[a].push_back(b);
      G[b].push_back(a);
    }
    if (k == 2)
      cout << n * (n - 1) / 2 << '\n';
    else {
      for (int i = 1; i <= n; i++) {
        vector<int> siema(n + 1);
        dfs(i);
      }
      cout << ans << '\n';
    }
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int test = 1;
    cin >> test;
    for (int i = 0; i < test; i++) solve();
  }
  