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;
  mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
  const long long maxn = 100 + 5, mod = 1e9 + 7, inf = mod;
  long long n, k;
  vector<long long> g[maxn];
  long long ans;
  long long cnt[maxn];
  vector<long long> dis[maxn];
  long long dp[maxn];
  void dfs(long long i, long long p, long long d) {
    cnt[d]++;
    for (long long j : g[i]) {
      if (j != p) {
        dfs(j, i, d + 1);
      }
    }
  }
  long long get(vector<long long>& v) {
    fill(begin(dp), end(dp), 0);
    dp[0] = 1;
    for (long long x : v) {
      for (long long num = (long long)n; num >= (long long)1; num--) {
        dp[num] = (dp[num] + dp[num - 1] * x) % mod;
      }
    }
    return dp[k];
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long tc;
    cin >> tc;
    while (tc--) {
      cin >> n >> k;
      ans = 0;
      fill(g + 1, g + n + 1, vector<long long>());
      for (long long i = (long long)1; i <= (long long)n - 1; i++) {
        long long u, v;
        cin >> u >> v;
        g[u].emplace_back(v);
        g[v].emplace_back(u);
      }
      if (k == 1) {
        cout << n << '\n';
        continue;
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        continue;
      }
      for (long long i = (long long)1; i <= (long long)n; i++) {
        for (long long d = (long long)0; d <= (long long)n; d++) dis[d].clear();
        for (long long j : g[i]) {
          fill(begin(cnt), end(cnt), 0);
          dfs(j, i, 1);
          for (long long d = (long long)1; d <= (long long)n; d++) {
            dis[d].emplace_back(cnt[d]);
          }
        }
        for (long long d = (long long)1; d <= (long long)n; d++) {
          ans += get(dis[d]);
          ans %= mod;
        }
      }
      cout << ans << '\n';
    }
  }
  