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;
  vector<vector<int> > ed;
  const int m = 1e9 + 7;
  void dfs(int cur, int pr, vector<int>& dist) {
    dist[cur] = dist[pr] + 1;
    for (auto to : ed[cur]) {
      if (to == pr) continue;
      dfs(to, cur, dist);
    }
  }
  void solve() {
    int n, k;
    cin >> n >> k;
    ed.clear();
    ed.resize(n + 1);
    for (int i = 0; i < n - 1; ++i) {
      int u, v;
      cin >> u >> v;
      ed[u].push_back(v);
      ed[v].push_back(u);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << "\n";
      return;
    }
    int ans = 0;
    for (int i = 1; i <= n; ++i) {
      vector<int> dist(n + 1);
      vector<vector<int> > arr(n + 1, vector<int>(n + 1));
      for (auto to : ed[i]) {
        dfs(to, i, dist);
        for (int j = 1; j <= n; ++j) {
          if (dist[j] == 0) continue;
          arr[to][dist[j]]++;
        }
        dist.assign(n + 1, 0);
      }
      for (int l = 1; l <= n; ++l) {
        vector<vector<int> > dp(n + 1, vector<int>(n + 1));
        int prev = 0;
        dp[0][0] = 1;
        for (auto to : ed[i]) {
          for (int j = 0; j <= n; ++j) {
            dp[to][j] = dp[prev][j];
            if (j != 0)
              dp[to][j] += (long long)dp[prev][j - 1] * arr[to][l] % m,
                  dp[to][j] %= m;
          }
          prev = to;
        }
        ans += dp[prev][k];
        ans %= m;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int tst;
    tst = 1;
    cin >> tst;
    while (tst--) {
      solve();
    }
    return 0;
  }
  