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;
  long long mod = 1e9 + 7;
  vector<vector<long long> > d(110);
  vector<bool> u(110);
  vector<long long> t;
  long long dfs(long long v, long long dist, long long pr) {
    u[v] = 1;
    if (!dist) {
      return 1;
    }
    long long m = 0;
    for (long long i = 0; i < d[v].size(); i++)
      if (!u[d[v][i]]) {
        long long b = dfs(d[v][i], dist - 1, pr);
        m += b;
        if (v == pr && b) t.push_back(b);
      }
    return m;
  }
  long long f(long long k) {
    long long m = mod - 2;
    long long a = 1;
    while (m) {
      if (m % 2) a = 1ll * a * k % mod;
      k = 1ll * k * k % mod;
      m /= 2;
    }
    return a;
  }
  void solve() {
    long long n, m, k, ans = 0;
    cin >> n >> k;
    for (long long i = 0; i < n; i++) d[i].clear();
    for (long long i = 0; i < n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      u--;
      v--;
      d[u].push_back(v);
      d[v].push_back(u);
    }
    if (k == 1) {
      cout << n << '\n';
      return;
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      return;
    }
    for (long long i = 0; i < n; i++) {
      for (long long j = 1; j < n; j++) {
        long long ans1 = 0;
        for (long long h = 0; h < n; h++) u[h] = 0;
        t.clear();
        m = dfs(i, j, i);
        if (t.size() >= k) {
          long long dp[110][110] = {0};
          for (long long j = 0; j < k; j++) {
            long long sum;
            if (!j)
              sum = 1;
            else
              sum = dp[j - 1][j - 1];
            for (long long i = j; i < t.size(); i++) {
              dp[i][j] = 1ll * sum * t[i] % mod;
              if (j) sum = (sum + dp[i][j - 1]) % mod;
            }
          }
          for (long long i = 0; i < t.size(); i++)
            ans1 = (ans1 + dp[i][k - 1]) % mod;
          ans = (ans + ans1) % mod;
        }
      }
    }
    cout << ans << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  