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;
  struct node {
    map<long long, long long> st;
    node() { st = {}; }
  };
  const long long maxn = 100;
  vector<long long> g[maxn];
  long long h[maxn];
  const long long mod = 1e9 + 7;
  vector<node *> all;
  node *dfs(long long v, long long p) {
    if (p == -1)
      h[v] = 0;
    else
      h[v] = h[p] + 1;
    vector<long long> ans;
    vector<node *> gg;
    for (long long i = 0; i < g[v].size(); ++i) {
      long long to = g[v][i];
      if (to == p) continue;
      gg.push_back(dfs(to, v));
    }
    if (h[v] != 0) {
      node *res = new node();
      long long id = 0;
      for (long long j = 0; j < gg.size(); ++j) {
        if (gg[id]->st.size() < gg[j]->st.size()) id = j;
      }
      if (gg.size() != 0) res = gg[id];
      res->st[(h[v])]++;
      for (long long k = 0; k < gg.size(); ++k) {
        if (k == id) continue;
        for (auto y : gg[k]->st) {
          res->st[(y).first] += y.second;
        }
      }
      if (h[v] == 1) all.push_back(res);
      return res;
    }
    return new node();
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie();
    cout.tie();
    long long q;
    cin >> q;
    while (q--) {
      long long n, k;
      cin >> n >> k;
      for (long long i = 0; i < n; ++i) {
        g[i].clear();
      }
      for (long long m = 0; m < n - 1; ++m) {
        long long a, b;
        cin >> a >> b;
        a--;
        b--;
        g[a].push_back(b);
        g[b].push_back(a);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 % mod << "\n";
        continue;
      }
      long long ans = 0;
      for (long long j = 0; j < n; ++j) {
        all.clear();
        dfs(j, -1);
        for (long long i = 1; i < n; ++i) {
          vector<long long> kol;
          for (long long l = 0; l < all.size(); ++l) {
            if (all[l]->st.count(i)) kol.push_back(all[l]->st[i]);
          }
          if (kol.size() < k) continue;
          long long dp[kol.size() + 1][k + 1];
          for (long long i1 = 0; i1 < kol.size() + 1; ++i1) {
            for (long long l = 0; l < k + 1; ++l) {
              dp[i1][l] = 0;
            }
          }
          dp[0][0] = 1;
          for (long long m = 1; m < kol.size() + 1; ++m) {
            dp[m][0] = 1;
            for (long long l = 1; l < k + 1; ++l) {
              dp[m][l] += dp[m - 1][l];
              dp[m][l] += dp[m - 1][l - 1] * kol[m - 1] % mod;
              dp[m][l] %= mod;
            }
          }
          ans = (ans + dp[kol.size()][k]) % mod;
        }
      }
      cout << ans << "\n";
    }
  }
  