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 Mod = (long long)1e9 + 7;
  const long long MAXFAC = (long long)2e6 + 10;
  long long fac[MAXFAC];
  inline long long pw(long long x, long long y) {
    long long res = 1;
    while (y) {
      if (y & 1ll) res = res * x % Mod;
      x = x * x % Mod;
      y /= 2;
    }
    return res;
  }
  void init_fac(long long maxn = MAXFAC) {
    fac[0] = 1;
    for (long long i = 1; i < maxn; ++i) {
      fac[i] = fac[i - 1] * i % Mod;
    }
  }
  inline long long inv(long long x) { return pw(x, Mod - 2); };
  inline long long choose(long long n, long long r) {
    return fac[n] * (inv(fac[n - r]) * inv(fac[r]) % Mod) % Mod;
  }
  void preproc() { init_fac(110); }
  void solve() {
    long long n, k;
    cin >> n >> k;
    vector<vector<long long>> g(n + 1);
    for (long long e = 1; e <= n - 1; ++e) {
      long long u, v;
      cin >> u >> v;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    if (k == 2) {
      cout << choose(n, 2) << '\n';
      return;
    }
    vector<vector<long long>> f(n + 1, vector<long long>(n + 1));
    vector<vector<long long>> h(n + 1, vector<long long>(n + 1));
    function<void(long long, long long)> dfs_f = [&](long long u, long long pr) {
      f[u][0] = 1;
      for (auto it = g[u].begin(); it != g[u].end(); ++it)
        if (*it == pr) {
          g[u].erase(it);
          break;
        }
      for (long long v : g[u])
        if (v != pr) {
          dfs_f(v, u);
          for (long long d = 1; d <= n; ++d) f[u][d] += f[v][d - 1];
        }
    };
    dfs_f(1, 0);
    function<void(long long)> dfs_g = [&](long long u) {
      h[u][0] = 1;
      long long childs = (long long)g[u].size();
      vector<vector<long long>> pref(childs, vector<long long>(n + 1));
      vector<vector<long long>> suff(childs, vector<long long>(n + 1));
      for (long long i = 0; i < childs; ++i) {
        long long vi = g[u][i];
        long long rev_i = childs - i - 1;
        long long rev_vi = g[u][rev_i];
        for (long long d = 0; d <= n; ++d) {
          pref[i][d] = f[vi][d] + (i - 1 >= 0 ? pref[i - 1][d] : 0ll);
          suff[rev_i][d] =
              f[rev_vi][d] + (rev_i + 1 <= childs - 1 ? suff[rev_i + 1][d] : 0ll);
        }
      }
      for (long long i = 0; i < childs; ++i) {
        long long vi = g[u][i];
        for (long long d = 1; d <= n; ++d) {
          h[vi][d] += h[u][d - 1];
          if (d - 2 >= 0 and i - 1 >= 0) h[vi][d] += pref[i - 1][d - 2];
          if (d - 2 >= 0 and i + 1 <= childs - 1) h[vi][d] += suff[i + 1][d - 2];
        }
      }
      for (long long v : g[u]) dfs_g(v);
    };
    dfs_g(1);
    long long res = 0;
    for (long long u = 1; u <= n; ++u) {
      long long neigh = (long long)g[u].size() + (u != 1);
      if (neigh < k) continue;
      for (long long d = 1; d <= n; ++d) {
        vector<long long> a;
        if (h[u][d] > 0) a.push_back(h[u][d]);
        for (long long v : g[u])
          if (f[v][d - 1] > 0) a.push_back(f[v][d - 1]);
        if ((long long)a.size() < k) break;
        long long sz = (long long)a.size();
        vector<vector<long long>> dp(sz + 1, vector<long long>(k + 1));
        for (long long i = 0; i <= sz; ++i) dp[i][0] = 1;
        for (long long vk = 1; vk <= k; ++vk)
          for (long long i = 1; i <= sz; ++i)
            dp[i][vk] = (dp[i - 1][vk] + dp[i - 1][vk - 1] * a[i - 1]) % Mod;
        res = (res + dp[sz][k]) % Mod;
      }
    }
    res = (res) % Mod;
    cout << res << '\n';
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.precision(10);
    cout << fixed;
    preproc();
    long long tc = 1;
    cin >> tc;
    for (long long Tt = 1; Tt <= tc; ++Tt) {
      solve();
    }
    return EXIT_SUCCESS;
  }
  