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 LL = long long;
  using namespace std;
  const int N = 100 + 1, mod = 1e9 + 7;
  int D[N], H[N][N];
  vector<int> G[N];
  void add(int& a, int b) { a = (a + b) % mod; }
  void bfs(int v, int h) {
    int q[N], l = 0, r = 0;
    q[r++] = v;
    D[v] = 1;
    H[h][1]++;
    while (l < r) {
      int u = q[l++];
      for (auto to : G[u]) {
        if (D[to] != -1) continue;
        q[r++] = to;
        D[to] = D[u] + 1;
        H[h][D[to]]++;
      }
    }
  }
  int cnt(int v, int k) {
    memset(D, -1, sizeof(D));
    ;
    memset(H, 0, sizeof(H));
    ;
    D[v] = 0;
    int m = G[v].size();
    if (m < k) return 0;
    for (int h = 0; h < m; ++h) {
      auto to = G[v][h];
      bfs(to, h);
    }
    int c = 0;
    for (int d = 1; d < N; ++d) {
      int dp[m + 1][k + 1];
      memset(dp, 0, sizeof(dp));
      ;
      dp[0][0] = 1;
      for (int h = 1; h <= m; ++h) {
        dp[h][0] = 1;
        for (int j = 1; j <= k; ++j) {
          add(dp[h][j], dp[h - 1][j]);
          add(dp[h][j], ((LL)dp[h - 1][j - 1] * H[h - 1][d]) % mod);
        }
      }
      add(c, dp[m][k]);
    }
    return c;
  }
  int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    ;
    while (t--) {
      int n, k;
      cin >> n >> k;
      for (int i = 1; i <= n; ++i) G[i].clear();
      for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << '\n';
        continue;
      }
      int res = 0;
      for (int i = 1; i <= n; ++i) add(res, cnt(i, k));
      cout << res << '\n';
    }
  }
  