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: md = int(1e9) + 7
  N = 110
  
  
  def add(x, y):
      x += y
      if x >= md:
          x -= md
      if x < 0:
          x += md
      return x
  
  
  def mul(x, y):
      x *= y
      x %= md
      return x
  
  
  depth = [0] * N
  cnt = [0]
  g = [[] for _ in range(N)]
  
  
  def Dfs(v, pr):
      for to in g[v]:
          if to != pr:
              depth[to] = depth[v] + 1
              cnt[depth[to]] += 1
              Dfs(to, v)
  
  
  def solve(n, k, v):
      dp = [[1] + k * [0] for _ in range(n)]
      depth[v] = 0
      res = 0
      for to in g[v]:
          global cnt
          cnt = [0] * n
          depth[to] = 1
          cnt[depth[to]] += 1
          Dfs(to, v)
          # print(cnt)
          for dep in range(n):
              for ct in reversed(range(k)):
                  dp[dep][ct + 1] = add(dp[dep][ct + 1],
                                        mul(dp[dep][ct], cnt[dep]))
          # print(n, k, v, cnt, dp)
      for i in range(n):
          res = add(res, dp[i][k])
      return res
  
  
  t = int(input())
  for _ in range(t):
      input()
      n, k = map(int, input().split())
      for i in range(n):
          g[i].clear()
      for i in range(n - 1):
          u, v = map(int, input().split())
          u -= 1
          v -= 1
          g[u].append(v)
          g[v].append(u)
  
      if k == 2:
          print(int((n * (n - 1) / 2) % md))
          continue
  
      ans = 0
      for i in range(n):
          ans = add(ans, solve(n, k, i))
      print(ans)
  