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 ll = long long;
  using pii = pair<int, int>;
  constexpr int N = 101;
  constexpr int M = 1e9 + 7;
  ll ans;
  int n, k;
  vector<int> edges[N];
  int depth[N][N];
  int arr[N];
  int dp[N][N];
  void dfs(int cur, int par) {
    memset(depth[cur], 0, sizeof(int) * n);
    depth[cur][0] = 1;
    for (int i = (0); i < (edges[cur].size()); ++i) {
      int nxt = edges[cur][i];
      if (nxt == par) continue;
      dfs(nxt, cur);
      for (int j = (1); j < (n); ++j) {
        depth[cur][j] += depth[nxt][j - 1];
      }
    }
  }
  void solve(int root) {
    if (edges[root].size() < k) return;
    dfs(root, -1);
    for (int dist = (0); dist < (n); ++dist) {
      for (int i = (0); i < (edges[root].size()); ++i) {
        int child = edges[root][i];
        int x = depth[child][dist];
        arr[i] = x;
      }
      memset(dp, 0, sizeof(int) * N * N);
      for (int i = (0); i < (edges[root].size()); ++i) {
        dp[i][0] = 1;
      }
      for (int i = (1); i <= (edges[root].size()); ++i) {
        for (int j = (1); j <= (min(i, k)); ++j) {
          dp[i][j] = (dp[i - 1][j] + (ll)arr[i - 1] * dp[i - 1][j - 1]) % M;
        }
      }
      ans += dp[edges[root].size()][k];
      ans %= M;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int tc;
    cin >> tc;
    while (tc--) {
      cin >> n >> k;
      memset(edges, 0, sizeof(vector<int>) * N);
      ans = 0;
      for (int i = (0); i < (n - 1); ++i) {
        int u, v;
        cin >> u >> v;
        u--;
        v--;
        edges[u].push_back(v);
        edges[v].push_back(u);
      }
      if (k == 2) {
        ans = n * (n - 1) / 2;
        ans %= M;
      } else {
        for (int i = (0); i < (n); ++i) {
          solve(i);
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  