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;
  long long n, k, nr[105][105], dp[105][105], t;
  vector<int> muchii[200005];
  bool use[200005];
  void dfs(int nod) {
    use[nod] = 1;
    nr[nod][0] = 1;
    for (auto i : muchii[nod])
      if (!use[i]) {
        dfs(i);
        for (int dist = 1; dist <= n; dist++)
          nr[nod][dist] = (nr[nod][dist] + nr[i][dist - 1]) % mod;
      }
  }
  void solve() {
    cin >> n >> k;
    for (int i = 1; i <= n; i++) muchii[i].clear();
    for (int i = 1; i < n; i++) {
      int a, b;
      cin >> a >> b;
      muchii[a].push_back(b);
      muchii[b].push_back(a);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      return;
    }
    long long ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        use[j] = 0;
        for (int dist = 0; dist <= n; dist++) nr[j][dist] = 0;
      }
      dfs(i);
      for (int dist = 0; dist <= n; dist++) {
        dp[0][0] = 1;
        for (int j = 0; j < muchii[i].size(); j++) {
          dp[j + 1][0] = 1;
          for (int x = 1; x <= k; x++)
            dp[j + 1][x] =
                (dp[j][x] + dp[j][x - 1] * nr[muchii[i][j]][dist]) % mod;
        }
        ans = (ans + dp[muchii[i].size()][k]) % mod;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    cin >> t;
    while (t--) solve();
    return 0;
  }
  