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 MOD = (1e9 + 7);
  const long long MAX_VALEURS = (101);
  vector<long long> Voisins[MAX_VALEURS];
  long long Occ[MAX_VALEURS];
  long long DP[MAX_VALEURS][MAX_VALEURS];
  long long nbNoeuds, taille;
  void Dfs(long long noeud, long long pere, long long dist) {
    Occ[dist]++;
    for (long long dest : Voisins[noeud]) {
      if (dest != pere) Dfs(dest, noeud, dist + 1);
    }
    return;
  }
  void Solve() {
    scanf("%lld %lld", &nbNoeuds, &taille);
    for (long long i = 0; i < nbNoeuds; i++) {
      Voisins[i].clear();
    }
    for (long long i = 1; i < nbNoeuds; i++) {
      long long a, b;
      scanf("%lld %lld", &a, &b);
      Voisins[--a].push_back(--b);
      Voisins[b].push_back(a);
    }
    if (taille == 2) {
      printf("%lld\n", (nbNoeuds * (nbNoeuds - 1)) / 2);
      return;
    }
    long long ans = 0;
    for (long long i = 0; i < nbNoeuds; i++) {
      for (long long j = 0; j < nbNoeuds; j++) {
        fill_n(DP[j], nbNoeuds, 0LL);
        DP[j][0] = 1;
      }
      for (long long dest : Voisins[i]) {
        fill_n(Occ, nbNoeuds, 0LL);
        Dfs(dest, i, 1);
        for (long long d = 0; d < nbNoeuds; d++) {
          for (long long j = nbNoeuds - 1; j > 0; j--) {
            DP[d][j] += DP[d][j - 1] * Occ[d];
            DP[d][j] %= MOD;
          }
        }
      }
      for (long long d = 0; d < nbNoeuds; d++) {
        ans += DP[d][taille];
      }
      ans %= MOD;
    }
    printf("%lld\n", ans);
    return;
  }
  signed main() {
    long long nbTests = 1;
    scanf("%lld", &nbTests);
    while (nbTests--) {
      Solve();
    }
    return 0;
  }
  