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 long long N = 105, MOD = 1e9 + 7;
  vector<vector<long long>> adj;
  map<long long, map<long long, long long>> cnt;
  long long dep[N], ans, dp[N];
  long long n, k;
  void dfs(long long u, long long p, long long root) {
    cnt[root][dep[u]]++;
    for (auto v : adj[u]) {
      if (v == p) continue;
      dep[v] = dep[u] + 1;
      dfs(v, u, root);
    }
  }
  void solve() {
    cin >> n >> k;
    adj.clear();
    adj.resize(n + 1);
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    ans = 0;
    if (k == 2) {
      ans = n * (n - 1) / 2;
    } else {
      for (long long i = 1; i <= n; i++) {
        cnt.clear();
        for (auto v : adj[i]) {
          dep[v] = 1;
          dfs(v, i, v);
        }
        for (long long j = 1; j <= 100; j++) {
          fill(dp, dp + N, 0);
          dp[0] = 1;
          for (auto v : adj[i]) {
            for (long long kk = 100; ~kk; kk--) {
              dp[kk + 1] = (dp[kk + 1] + dp[kk] * cnt[v][j] % MOD);
            }
          }
          ans = (ans + dp[k]) % MOD;
        }
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
  }
  