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 double PI = acos(-1);
  const int INF = INT_MAX;
  const long long LLINF = 1000000000000000000LL;
  const long long MAX = 105;
  const long long MOD = 1000000007;
  long long N, K, deg;
  long long tree[MAX][MAX];
  long long cache[MAX][MAX];
  vector<int> adj[MAX];
  bool visited[MAX];
  void sub(long long i, long long d, long long idx) {
    tree[i][d]++;
    visited[idx] = true;
    for (auto& it : adj[idx]) {
      if (!visited[it]) {
        sub(i, d + 1, it);
      }
    }
  }
  long long DP(long long n, long long k, long long d) {
    if (k == 0) {
      return 1;
    }
    if (n == deg) {
      return 0;
    }
    long long& ret = cache[n][k];
    if (ret != -1) {
      return ret;
    }
    ret = DP(n + 1, k, d) + DP(n + 1, k - 1, d) * tree[n][d];
    ret %= MOD;
    return ret;
  }
  long long root(long long n) {
    deg = adj[n].size();
    visited[n] = true;
    int idx = 0;
    for (auto& it : adj[n]) {
      sub(idx, 0, it);
      ++idx;
    }
    long long ans = 0;
    for (int d = 0; d <= N; d++) {
      memset(cache, -1, sizeof(cache));
      ans += DP(0, K, d);
      ans %= MOD;
    }
    return ans;
  }
  void solve() {
    cin >> N >> K;
    for (int i = 0; i < N; i++) {
      adj[i].clear();
    }
    for (int i = 0; i < N - 1; i++) {
      long long a, b;
      cin >> a >> b;
      --a;
      --b;
      adj[a].push_back(b);
      adj[b].push_back(a);
    }
    if (K == 2) {
      cout << N * (N - 1) / 2 << "\n";
      return;
    }
    long long ret = 0;
    for (int i = 0; i < N; i++) {
      if (adj[i].size() >= K) {
        memset(tree, 0, sizeof(tree));
        memset(visited, 0, sizeof(visited));
        ret += root(i);
        ret %= MOD;
      }
    }
    cout << ret << "\n";
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ;
    int tc;
    cin >> tc;
    while (tc--) {
      solve();
    }
    return 0;
  }
  