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 int maxn = 2e5 + 7, mod = 1e9 + 7;
  long long dp[105][105], ans;
  vector<int> G[105], d;
  int n, k, mx;
  void dfs(int x, int f, int dep) {
    ++d[dep];
    if (dep > mx) mx = dep;
    for (int i : G[x]) {
      if (i == f) continue;
      dfs(i, x, dep + 1);
    }
  }
  inline void solve() {
    cin >> n >> k;
    ans = 0;
    for (int i = 1; i <= n; ++i) G[i].clear();
    for (int i = 2, x, y; i <= n; ++i) {
      cin >> x >> y;
      G[x].emplace_back(y);
      G[y].emplace_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << '\n';
      return;
    }
    vector<int> t(n + 1);
    for (int root = 1; root <= n; ++root) {
      vector<vector<int> > v;
      mx = 0;
      for (int x : G[root]) {
        d = t;
        dfs(x, root, 1);
        v.emplace_back(d);
      }
      int m = v.size();
      if (m < k) continue;
      for (int i = 1; i <= mx; ++i) {
        vector<int> p;
        int q = 0;
        for (auto &x : v) p.emplace_back(x[i]), q |= x[i];
        if (!q) break;
        for (int j = 1; j <= m; ++j)
          for (int x = 0; x <= k; ++x) dp[j][x] = 0;
        dp[0][0] = 1;
        for (int j = 1; j <= m; ++j) {
          for (int x = 0; x <= k; ++x) dp[j][x] = dp[j - 1][x];
          for (int x = 1; x <= k; ++x)
            dp[j][x] = (dp[j][x] + dp[j - 1][x - 1] * p[j - 1]) % mod;
        }
        ans = (ans + dp[m][k]) % mod;
      }
    }
    cout << ans << '\n';
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int _ = 1;
    cin >> _;
    while (_--) solve();
    return 0;
  }
  