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 mod1 = 998244353;
  const long long mod2 = 1000000007;
  int t, n, k, u, v;
  vector<int> con[105];
  int dp[105][105];
  int cnt[105][105];
  int par[105];
  int res, mx;
  int dfs(int i, int p = -1) {
    par[i] = p;
    int ret = -1;
    for (int j = 0, l = con[i].size(); j < l; ++j)
      if (con[i][j] != p) ret = max(ret, dfs(con[i][j], i));
    return ++ret;
  }
  int cnts(int i, int d) {
    if (cnt[i][d] != -1) return cnt[i][d];
    if (d == 0) return cnt[i][d] = 1;
    int ret = 0;
    for (int j = 0, l = con[i].size(); j < l; ++j)
      if (con[i][j] != par[i]) ret += cnts(con[i][j], d - 1);
    return cnt[i][d] = ret;
  }
  void Q() {
    cin >> n >> k;
    res = 0;
    for (int i = 1; i <= (n); ++i) con[i].clear();
    for (int i = 1; i < n; ++i) {
      cin >> u >> v;
      con[u].push_back(v);
      con[v].push_back(u);
    }
    if (k == 2) {
      cout << 1ll * (n - 1) * n / 2 % mod2 << endl;
      return;
    }
    for (int r = 1; r <= n; ++r) {
      memset(cnt, -1, sizeof(cnt));
      mx = dfs(r);
      for (int d = 1; d <= mx; ++d) {
        memset(dp, 0, sizeof(dp));
        dp[0][0] = 1;
        for (int i = 0, l = con[r].size(); i <= l; ++i)
          for (int j = 0; j <= i && j <= k; ++j) {
            dp[i + 1][j] += dp[i][j];
            if (dp[i + 1][j] >= mod2) dp[i + 1][j] -= mod2;
            if (i != l) {
              dp[i + 1][j + 1] += 1ll * dp[i][j] * cnts(con[r][i], d - 1) % mod2;
              if (dp[i + 1][j + 1] >= mod2) dp[i + 1][j + 1] -= mod2;
            }
          }
        res += dp[con[r].size()][k];
        if (res >= mod2) res -= mod2;
      }
    }
    cout << res << endl;
  }
  int main() {
    cin >> t;
    while (t--) Q();
    return 0;
  }
  