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 mod = 1e9 + 7;
  int t;
  int n, k;
  vector<int> g[110];
  int dep[110], dp[110][110];
  inline void upd(int &x, int y) { x = (x + y) % mod; }
  void dfs(int x, int fa, int d, int &cnt) {
    if (dep[x] == d) cnt++;
    for (int i = 0; i < (int)g[x].size(); i++) {
      int to = g[x][i];
      if (to == fa) continue;
      dep[to] = dep[x] + 1;
      dfs(to, x, d, cnt);
    }
  }
  void solve() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) g[i].clear();
    for (int i = 1; i < n; i++) {
      int u, v;
      cin >> u >> v;
      u--;
      v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << 1ll * n * (n - 1) / 2 % mod << endl;
      return;
    }
    int ans = 0;
    for (int r = 0; r < n; r++)
      for (int d = 1; d <= n; d++) {
        if ((int)g[r].size() < k) continue;
        dp[0][0] = 1;
        for (int i = 0; i < (int)g[r].size(); i++) {
          int cnt = 0;
          dep[g[r][i]] = 1;
          dfs(g[r][i], r, d, cnt);
          for (int j = 0; j <= i + 1; j++) dp[i + 1][j] = 0;
          for (int j = 0; j <= i; j++) {
            if (cnt > 0) upd(dp[i + 1][j + 1], 1ll * dp[i][j] * cnt % mod);
            upd(dp[i + 1][j], dp[i][j]);
          }
        }
        cout << endl;
        upd(ans, dp[(int)g[r].size()][k]);
      }
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  