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 = 1e9 + 7;
  const long long Fact_Length = 1e5 + 1;
  long long Power(long long base, long long expo) {
    long long $result = 1;
    base %= mod;
    while (expo) {
      if (expo % 2 == 1) $result = ($result * base) % mod;
      base = (base * base) % mod;
      expo /= 2;
    }
    return $result;
  }
  long long Mod_Inv(long long $a) { return Power($a, mod - 2); }
  long long Factorial[Fact_Length];
  long long Make_Factorial() {
    Factorial[0] = 1;
    for (long long i = 1; i < Fact_Length; i++) {
      Factorial[i] = (i * Factorial[i - 1]) % mod;
    }
    return 0;
  }
  long long Implement_Make_Factorial = Make_Factorial();
  long long nCr(long long $n, long long $r) {
    if ($n < $r || $n < 0 || $r < 0) return 0;
    long long $ans = (Factorial[$n] * Mod_Inv(Factorial[$r])) % mod;
    $ans = ($ans * Mod_Inv(Factorial[$n - $r])) % mod;
    return $ans;
  }
  const long long N = 101;
  vector<long long> adj[N];
  long long n, k;
  long long dep[N];
  void DFS(long long chi, long long par, long long cur) {
    dep[cur]++;
    for (auto x : adj[chi]) {
      if (x == par) continue;
      DFS(x, chi, cur + 1);
    }
  }
  long long calc(long long b[], long long m) {
    if (m < k) return 0;
    long long dp[m + 1][k + 1];
    memset(dp, 0, sizeof(dp));
    for (long long i = 0; i <= m; i++) dp[i][0] = 1;
    for (long long i = 1; i <= m; i++) {
      for (long long j = 1; j <= k; j++) {
        dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] * b[i];
        dp[i][j] %= mod;
      }
    }
    return dp[m][k];
  }
  long long val(long long root) {
    vector<long long> a[n + 1];
    long long m = adj[root].size();
    for (auto x : adj[root]) {
      for (long long i = 1; i <= n; i++) dep[i] = 0;
      DFS(x, root, 1);
      for (long long i = 1; i <= n; i++) a[i].push_back(dep[i]);
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      sort(a[i].begin(), a[i].end());
      long long b[m + 1];
      for (long long j = 1; j <= m; j++) b[j] = a[i][j - 1];
      ans += calc(b, m);
      ans %= mod;
    }
    return ans;
  }
  void solve() {
    cin >> n >> k;
    for (long long i = 1; i <= n; i++) adj[i].clear();
    for (long long i = 1; i <= n - 1; i++) {
      long long u, v;
      cin >> u >> v;
      adj[u].push_back(v);
      adj[v].push_back(u);
    }
    if (k == 2) {
      long long ans = (n * (n - 1)) / 2;
      cout << ans << endl;
      return;
    }
    long long ans = 0;
    for (long long i = 1; i <= n; i++) {
      ans += val(i);
      ans %= mod;
    }
    cout << ans << endl;
  }
  int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    long long t;
    cin >> t;
    while (t--) solve();
    return 0;
  }
  