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 int mod = 1e9 + 7;
  const int M = 105;
  long long int n, k;
  vector<long long int> a[M];
  long long int d[M][M];
  long long int child;
  void dfs(long long int v, long long int p, long long int dep) {
    d[child][dep]++;
    for (long long int u : a[v]) {
      if (u == p) continue;
      dfs(u, v, dep + 1);
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    long long int TT = clock();
    long long int t;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (long long int i = 0; i < n; i++) a[i].clear();
      for (long long int i = 0; i < n - 1; i++) {
        long long int u, v;
        cin >> u >> v;
        u--, v--;
        a[u].push_back(v);
        a[v].push_back(u);
      }
      if (k == 2) {
        long long int ans = n * (n - 1) / 2;
        ans %= mod;
        cout << ans << "\n";
        continue;
      }
      long long int ans = 0;
      for (long long int i = 0; i < n; i++) {
        memset(d, 0, sizeof(d));
        for (long long int j = 0; j < a[i].size(); j++) {
          child = j;
          dfs(a[i][j], i, 1);
        }
        for (long long int dep = 1; dep <= n; dep++) {
          vector<long long int> c;
          for (long long int j = 0; j < n; j++)
            if (d[j][dep] != 0) c.push_back(d[j][dep]);
          long long int r = c.size();
          if (r < k) continue;
          vector<vector<long long int>> dp(r + 1, vector<long long int>(k + 2));
          dp[r][k] = 1;
          for (long long int i = r - 1; i >= 0; i--) {
            for (long long int j = k; j >= 0; j--) {
              dp[i][j] = (dp[i + 1][j] + dp[i + 1][j + 1] * c[i] % mod) % mod;
            }
          }
          ans = (ans + dp[0][0]) % mod;
        }
      }
      cout << ans << "\n";
    }
    cerr << "\n\nTIME: " << (long double)(clock() - TT) / CLOCKS_PER_SEC
         << " sec\n";
    TT = clock();
    return 0;
  }
  