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<int> g[101];
  vector<long long> v[101];
  bool check[101];
  long long dp[101][101];
  void go(int root, int parent, int k) {
    if (!check[k]) {
      v[k].push_back(1);
      check[k] = true;
    } else
      v[k].back()++;
    for (int i = 0; i < (int)g[root].size(); i++)
      if (g[root][i] != parent) go(g[root][i], root, k + 1);
  }
  long long f(int k, int m, int t) {
    if (k < 0) return m == 0 ? 1ll : 0ll;
    if (dp[k][m] != -1) return dp[k][m];
    long long temp = f(k - 1, m, t);
    if (m > 0) temp = (temp + f(k - 1, m - 1, t) * v[t][k]) % 1000000007;
    dp[k][m] = temp;
    return temp;
  }
  int main() {
    int t, n, k, a, b;
    long long ans, temp;
    scanf("%d", &t);
    for (; t > 0; t--) {
      scanf("%d %d", &n, &k);
      for (int i = 1; i < n; i++) {
        scanf("%d %d", &a, &b);
        g[a].push_back(b);
        g[b].push_back(a);
      }
      if (k == 2)
        printf("%d\n", n * (n - 1) >> 1);
      else {
        ans = 0;
        for (int i = 1; i <= n; i++) {
          if ((int)g[i].size() < k) continue;
          for (int j = 0; j < (int)g[i].size(); j++) {
            memset(check, false, sizeof(check));
            go(g[i][j], i, 1);
          }
          for (int j = 1; j <= n; j++) {
            memset(dp, -1, sizeof(dp));
            temp = f((int)v[j].size() - 1, k, j);
            ans = (ans + temp) % 1000000007;
            v[j].clear();
          }
        }
        printf("%lld\n", ans);
      }
      for (int i = 1; i <= n; i++) g[i].clear();
    }
    return 0;
  }
  