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 int N = 110;
  int dp[N][N];
  int cnt[N];
  int dep[N];
  vector<int> v[N];
  const int mod = 1e9 + 7;
  void dfs(int u, int fa, int d) {
    cnt[d]++;
    for (auto j : v[u]) {
      if (j == fa) continue;
      dfs(j, u, d + 1);
    }
  }
  void solve() {
    memset(dp, 0, sizeof dp);
    int n, m;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) v[i].clear();
    for (int i = 0; i < n - 1; i++) {
      int a, b;
      cin >> a >> b;
      v[a].push_back(b);
      v[b].push_back(a);
    }
    if (m == 2) {
      cout << n * (n - 1) / 2 << endl;
      return;
    }
    int ans = 0;
    for (int root = 1; root <= n; root++) {
      int ct = 0;
      for (int dd = 1; dd <= n; dd++) {
        dp[root][0] = 1;
        for (auto j : v[root]) {
          for (int kk = 1; kk <= n; kk++) cnt[kk] = 0;
          dfs(j, root, 1);
          if (cnt[dd])
            for (int k = m; k >= 1; k--)
              dp[root][k] =
                  (1ll * cnt[dd] * dp[root][k - 1] % mod + dp[root][k]) % mod;
        }
        ans = (ans + dp[root][m]) % mod;
        for (int j = 1; j <= m; j++) dp[root][j] = 0;
      }
    }
    cout << ans << endl;
  }
  int main() {
    ios::sync_with_stdio(false);
    int t;
    cin >> t;
    while (t--) {
      solve();
    }
    return 0;
  }
  