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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimization("unroll-loops")
  const int N = 2e5 + 5;
  const long long int mod = 1e9 + 7;
  const long long int Mod = 998244353;
  const long double Pi = acos(-1);
  const long long int Inf = 4e18;
  const long double Eps = 1e-9;
  int dx[9] = {0, 1, -1, 0, 0, 1, 1, -1, -1};
  int dy[9] = {0, 0, 0, 1, -1, 1, -1, 1, -1};
  using namespace std;
  long long int d[101][101];
  vector<long long int> g[101];
  long long int cnt;
  void Dfs(long long int t, long long int p, long long int k, long long int src) {
    cnt += (d[src][t] == k);
    for (auto u : g[t]) {
      if (u != p) {
        d[src][u] = d[src][t] + 1;
        Dfs(u, t, k, src);
      }
    }
  }
  void TestCase() {
    long long int n, k, ans = 0;
    cin >> n >> k;
    for (long long int i = 1, u, v; i < n; i++) {
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    memset(d, 0, sizeof(d));
    if (k == 2)
      cout << (n * (n - 1)) / 2;
    else {
      vector<long long int> take;
      vector<vector<long long int> > dp;
      for (long long int c = 2; c <= 100; c += 2) {
        for (long long int i = 1; i <= n; i++) {
          for (auto x : g[i]) {
            cnt = 0;
            d[i][x] = 1;
            Dfs(x, i, c / 2, i);
            take.push_back(cnt);
          }
          long long int m = (long long int)take.size();
          dp.resize(m + 1);
          for (long long int j = 0; j <= m; j++) dp[j].assign(k + 1, 0);
          dp[0][0] = 1;
          for (long long int j = 1; j <= m; j++) {
            for (long long int u = 0; u <= k; u++) dp[j][u] = dp[j - 1][u];
            for (long long int u = 1; u <= k; u++) {
              dp[j][u] += (dp[j - 1][u - 1] * take[j - 1]) % mod;
              dp[j][u] %= mod;
            }
          }
          ans += dp[m][k];
          ans %= mod;
          take.clear();
        }
      }
      cout << ans;
    }
    for (long long int i = 1; i <= n; i++) g[i].clear();
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    ;
    int T = 1;
    cin >> T;
    while (T--) {
      TestCase();
      cout << "\n";
    }
    return 0;
  }
  