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
  t=int(input())
  mod=10**9+7
  for _ in range(t):
      tmp=input()
      n,k=map(int,input().split())
      g=[[] for i in range(n)]
      for __ in range(n-1):
          u,v=map(int,input().split())
          g[u-1].append(v-1)
          g[v-1].append(u-1)
      f=[1]*101
      ans=0
      def dfs(u,depth):
          global ans
          f[u]=0
          for v in g[u]:
              if f[v]:
                  dfs(v,depth+1)
                  for j in range(n):
                      dp[u][j]+=dp[v][j]
          dp[u][depth]+=1
          if depth==0:
              for i in range(n):
                  c=[0]*(k+1)
                  c[0]=1
                  for to in g[u]:
                      if f[to] and dp[to][i]:
                          for kk in range(1,k+1)[::-1]:
                              c[kk]+=c[kk-1]*dp[to][i]%mod
                              c[kk]%=mod
                  ans=(ans+c[k])%mod
          f[u]=1
      for i in range(n):
          dp=[[0]*n for i in range(n)]
          dfs(i,0)
      if k==2:
          print(n*(n-1)//2)
      else:
          print(ans%mod)