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: for _ in range(int(input())):
      c = input()
      n, K = map(int, input().split())
      e = [[] for i in range(n)]
      for i in range(n-1):
          t1, t2 = map(int, input().split())
          t1 -= 1
          t2 -= 1
          e[t1].append(t2)
          e[t2].append(t1)
          
      bj = [1]*101
  
      ans = [0]
      
      def dfs(x, depth):
          bj[x] = 0
          for i in e[x]:
              if bj[i]:
                  dfs(i, depth+1)
                  for j in range(n):
                      a[x][j] += a[i][j]
          a[x][depth] += 1
  
          if depth == 0:
              for i in range(n):
                  f = [0]*(K+1)
                  f[0] = 1
                  for j in e[x]:
  
                      if bj[j] and a[j][i]:
                          for k in range(K, 0, -1):
                              f[k] = (f[k] + f[k-1]*a[j][i]) % 1000000007
                  ans[0] = (ans[0] + f[K]) % 1000000007
          bj[x] = 1
  
  
      for i in range(n):
          a = [[0] * n for j in range(n)]
          dfs(i, 0)
      if K == 2:
          print(n*(n-1)//2)
      else:
          print(ans[0]%1000000007)
  