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: import sys
  input = sys.stdin.readline
  from collections import defaultdict as dd
  
  MOD = 10 ** 9 + 7
  
  def main():
      input()
      n, k = map(int, input().split())
      edges = [[] for _ in range(n)]
      for _ in range(n - 1):
          u, v = map(int, input().split())
          u -= 1
          v -= 1
          edges[u].append(v)
          edges[v].append(u)
      if k == 2:
          print(n * (n - 1) // 2)
          return
      
      global ans
      ans = 0
      def dfs(pos, bpos):
          ret = dd(int)
          ret[0] = 1
          if bpos == -1:
              lst = dd(list)
          for npos in edges[pos]:
              if npos == bpos:
                  continue
              x = dfs(npos, pos)
              for k2, v in x.items():
                  k2 += 1
                  ret[k2] += v
                  if bpos == -1:
                      lst[k2].append(v)
          if bpos == -1:
              return lst
          else:
              return ret
      for i in range(n):
          lst = dfs(i, -1)
          for k2, v in lst.items():
              if len(v) < k:
                  continue
              dp = [0] * (k + 1)
              dp[0] = 1
              for i in v:
                  for j in range(k, 0, -1):
                      dp[j] += dp[j - 1] * i
                      dp[j] %= MOD
              ans += dp[k]
              ans %= MOD
      print(ans % MOD)
      
  for _ in range(int(input())):
      main()
  