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;
  int tcase, n, K;
  long long ans;
  vector<int> e[110];
  int cnt[110][110];
  long long f[110];
  void dfs(int sn, int fa, int r, int d) {
    cnt[r][d]++;
    for (int fn : e[sn])
      if (fn != fa) dfs(fn, sn, r, d + 1);
  }
  long long gao(int r) {
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) cnt[i][j] = 0;
    for (int fn : e[r]) dfs(fn, r, fn, 1);
    long long ret = 0;
    for (int t = 1; t <= n; t++) {
      for (int i = 0; i <= n; i++) f[i] = 0;
      f[0] = 1;
      int lim = 0;
      for (int fn : e[r]) {
        lim++;
        for (int i = min(lim, K); i; i--)
          f[i] = (f[i] + f[i - 1] * cnt[fn][t]) % 1000000007;
      }
      ret = (ret + f[K]) % 1000000007;
    }
    return ret;
  }
  int main() {
    scanf("%d", &tcase);
    while (tcase--) {
      scanf("%d%d", &n, &K);
      for (int i = 1; i <= n; i++) e[i].clear();
      for (int i = 1; i < n; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        e[a].push_back(b);
        e[b].push_back(a);
      }
      if (K == 2) {
        printf("%d\n", n * (n - 1) / 2);
        continue;
      }
      ans = 0;
      for (int i = 1; i <= n; i++) ans = (ans + gao(i)) % 1000000007;
      printf("%lld\n", ans);
    }
    return 0;
  }
  