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 long long maxn = 1e2 + 5;
  const long long mod = 1e9 + 7;
  const long long INF = 1e9 + 9;
  vector<long long> g[maxn];
  long long dp[maxn][maxn];
  long long depth[maxn][maxn];
  long long maxdep, cc;
  void dfs(long long x, long long p, long long t, long long level) {
    maxdep = max(maxdep, level);
    if (t == -1) {
      long long tot = 0;
      for (auto i : g[x]) dfs(i, x, ++tot, 1);
      cc = tot;
    } else {
      depth[t][level]++;
      for (auto i : g[x]) {
        if (i == p) continue;
        dfs(i, x, t, level + 1);
      }
    }
  }
  long long rundp(long long n, long long level, long long k) {
    dp[0][0] = 1;
    for (long long i = 1; i <= n; i++) {
      for (long long c = 0; c <= cc; c++) {
        if (c == 0)
          dp[i][c] = dp[i - 1][c];
        else
          dp[i][c] =
              (dp[i - 1][c] + dp[i - 1][c - 1] * depth[i][level] % mod) % mod;
      }
    }
    return dp[cc][k];
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    long long t;
    cin >> t;
    while (t--) {
      long long n, k, ans = 0;
      cin >> n >> k;
      for (long long i = 1; i <= n; i++) g[i].clear();
      for (long long i = 1; i < n; i++) {
        long long 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;
      }
      for (long long i = 1; i <= n; i++) {
        maxdep = 0;
        memset(depth, 0, sizeof(depth));
        dfs(i, -1, -1, 0);
        for (long long j = 1; j <= maxdep; j++) {
          memset(dp, 0, sizeof(dp));
          ans = (rundp(n, j, k) + ans) % mod;
        }
      }
      cout << ans << '\n';
    }
  }
  