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>
  std::vector<int> V[110];
  int count[110][110][110];
  void func(int k, int prev, int top1, int top2, int height) {
    count[top1][top2][height]++;
    for (int i = 0; i < V[k].size(); i++) {
      if (V[k][i] == prev) continue;
      func(V[k][i], k, top1, top2, height + 1);
    }
  }
  int x[110];
  long long int DP[110][110];
  int main() {
    int T;
    scanf("%d", &T);
    while (T--) {
      int a, b;
      scanf("%d%d", &a, &b);
      for (int i = 1; i <= a; i++) V[i].clear();
      for (int i = 1; i <= a; i++)
        for (int j = 1; j <= a; j++)
          for (int k = 0; k <= a; k++) count[i][j][k] = 0;
      for (int i = 1; i < a; i++) {
        int c, d;
        scanf("%d%d", &c, &d);
        V[c].push_back(d);
        V[d].push_back(c);
      }
      if (b == 2) {
        printf("%d\n", a * (a - 1) / 2);
        continue;
      }
      for (int i = 1; i <= a; i++)
        for (int j = 0; j < V[i].size(); j++) func(V[i][j], i, V[i][j], i, 1);
      long long int ans = 0;
      for (int i = 1; i <= a; i++) {
        if (V[i].size() >= b) {
          for (int j = 2; j <= a; j += 2) {
            int t = V[i].size();
            for (int k = 0; k < t; k++) x[k] = count[V[i][k]][i][j / 2];
            for (int k = 1; k <= b; k++) DP[t][k] = 0;
            DP[t][0] = 1;
            for (int l = t - 1; l >= 0; l--) {
              for (int k = 0; k <= b; k++) DP[l][k] = DP[l + 1][k];
              for (int k = 1; k <= b; k++)
                DP[l][k] += DP[l + 1][k - 1] * x[l], DP[l][k] %= 1000000007;
            }
            ans += DP[0][b], ans %= 1000000007;
          }
        }
      }
      printf("%lld\n", ans);
    }
  }
  