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 deque, Counter
  
  mod = 10 ** 9 + 7
  for _ in range(int(input())):
      input()
      n, k = map(int, input().split())
      G = [[] for _ in range(n)]
      for _ in range(n - 1):
          a, b = map(int, input().split())
          a -= 1
          b -= 1
          G[a].append(b)
          G[b].append(a)
  
      if k == 2:
          print(n * (n - 1) // 2 % mod)
          continue
  
      ans = 0
      for r in range(n):
          cnt = [0] * n
          dp = [[0] * (k + 1) for _ in range(n)]
          for i in range(1, n):
              dp[i][0] = 1
  
          def dfs(i, par, d):
              cnt[d] += 1
              for j in G[i]:
                  if j == par: continue
                  dfs(j, i, d + 1)
  
          for x in G[r]:
              cnt = [0] * n
              dfs(x, r, 1)
              for i in range(1, n):
                  if cnt[i] == 0: break
                  for j in range(k, 0, -1):
                      dp[i][j] += dp[i][j - 1] * cnt[i]
                      dp[i][j] %= mod
  
          for i in range(1, n):
              ans += dp[i][k]
              ans %= mod
  
      print(ans)
  