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 M = 1e2 + 5;
  const int mod = 1e9 + 7;
  long long dp[M][M];
  int k, n, freq[M][M];
  vector<int> adj[M], dpV;
  void DFS(int x, int p) {
    int i;
    freq[x][0] = 1;
    for (int y : adj[x]) {
      if (p ^ y) {
        DFS(y, x);
        for (i = 1; i < n; i++) freq[x][i] += freq[y][i - 1];
      }
    }
  }
  long long DP(int cur, int cnt) {
    if (cnt > k) return 0;
    if (cur == (int)dpV.size()) return (cnt == k);
    long long &ans = dp[cur][cnt];
    if (ans > -1) return ans;
    ans = DP(cur + 1, cnt);
    ans += dpV[cur] * DP(cur + 1, cnt + 1);
    return ans %= mod;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long ans;
    int i, j, t, x, y;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (i = 1; i <= n; i++) adj[i].clear();
      for (i = 1; i < n; i++) {
        cin >> x >> y;
        adj[x].emplace_back(y);
        adj[y].emplace_back(x);
      }
      if (k == 2)
        ans = (1ll * n * (n - 1)) / 2;
      else {
        ans = 0;
        for (i = 1; i <= n; i++) {
          memset(freq, 0, sizeof(freq));
          DFS(i, 0);
          for (j = 0; j < n - 1; j++) {
            dpV.clear();
            for (int it : adj[i]) {
              if (freq[it][j]) dpV.emplace_back(freq[it][j]);
            }
            if ((int)dpV.size() >= k) {
              memset(dp, -1, sizeof(dp));
              ans += DP(0, 0);
              if (ans >= mod) ans -= mod;
            }
          }
        }
      }
      cout << ans << '\n';
    }
    return 0;
  }
  