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: from collections import defaultdict
  
  def memoize(function):
      cache = {}
      def decorated_function(*args):
          try:
              return cache[args]
          except KeyError:
              val = function(*args)
              cache[args] = val
              return val
      return decorated_function
  
  @memoize
  def inverseMod(n, mod):
      if 1 == n: return 1
  
      return ((mod - mod//n) * inverseMod(mod % n, mod)) % mod
  
  M = 10**9 + 7
  
  F = [1, 1]
  for n in range(2, 10**5 + 1):
      F.append((F[-1] * n) % M)
  
  RF = [1, 1]
  for n in range(2, 10**5 + 1):
      RF.append((RF[-1] * inverseMod(n, M)) % M)
  
  def C(n, k):
      return (F[n] * RF[n - k] * RF[k]) % M
  
  def count(Q, k):
      size = len(Q)
      dp = [[0] * size for _ in range(k + 1)]
      for used in range(size):
          dp[0][used] = 1
  
      for taken in range(1, k + 1):
          for used in range(taken - 1, size):
              dp[taken][used] = (dp[taken][used - 1] + dp[taken - 1][used - 1]*len(Q[used])) % M
              
  #    for d in dp:
  #        print(d)
      return dp[k][size - 1]
  
  #print("!", count([[14], [51], [17]], 3))
  
  #exit(2)
  
  T = int(input())
  
  for _ in range(T):
      input()
      n, k = map(int, input().split(' '))
  
      G = defaultdict(list)
      for _ in range(n - 1):
          s, e = map(int, input().split(' '))
          G[s].append(e)
          G[e].append(s)
  
      if k == 2:
          r = C(n, k)
      else:
          r = 0
          for s in range(1, n + 1):
              Q = [[g] for g in G[s]]
              r += count(Q, k)
  
              V = {s}
              while len(Q) > 0:
                  nQ = []
                  for sq in Q:
                      nsq = []
                      for e in sq:
                          V.add(e)
                          for g in G[e]:
                              if g not in V:
                                  nsq.append(g)
                      if len(nsq) > 0:
                          nQ.append(nsq)
  
                  if len(nQ) >= k:
                      r += count(nQ, k)
                  if len(nQ) <= 1:
                      break
                  Q = nQ
  
      print(r % M)