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;
  const int mxN = 1e3 + 5, mod = 1e9 + 7, LOG = 20;
  int n, k, cnt[mxN], ans;
  int dp[mxN][mxN];
  vector<int> adj[mxN];
  void calc(int v, int p, int dep = 1) {
    ++cnt[dep];
    for (auto to : adj[v]) {
      if (to != p) calc(to, v, dep + 1);
    }
  }
  void get(int node) {
    for (int i = 0; i <= n; ++i) {
      for (int j = 0; j <= n; ++j) dp[i][j] = 0;
    }
    for (int i = 1; i <= n; ++i) dp[0][i] = 1;
    int now = 0;
    for (auto to : adj[node]) {
      calc(to, node);
      ++now;
      for (int d = 1; d <= n; ++d) {
        for (int j = now; j > 0; j--)
          dp[j][d] = (dp[j][d] + dp[j - 1][d] * 1ll * cnt[d] % mod) % mod;
      }
      for (int i = 1; i <= n; ++i) cnt[i] = 0;
    }
    for (int i = 1; i <= n; ++i) ans = (ans + dp[k][i]) % mod;
  }
  void solve() {
    cin >> n >> k;
    ans = 0;
    for (int i = 1; i <= n; ++i) adj[i].clear();
    for (int i = 1, u, v; i < n; ++i) {
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      cout << (n * 1ll * (n - 1) / 2) % mod;
      return;
    }
    for (int c = 1; c <= n; ++c) {
      get(c);
    }
    cout << ans;
  }
  int main() {
    ios ::sync_with_stdio(false), cin.tie(nullptr);
    int t;
    cin >> t;
    while (t--) {
      solve();
      cout << '\n';
    }
    return 0;
  }
  