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: #include <bits/stdc++.h>
  using namespace std;
  template <typename T, typename T1>
  T amax(T &a, T1 b) {
    if (b > a) a = b;
    return a;
  }
  template <typename T, typename T1>
  T amin(T &a, T1 b) {
    if (b < a) a = b;
    return a;
  }
  const long long N = 1e2 + 5;
  long long freq[N], n, k;
  vector<long long> v[N];
  void dfs(long long z, long long p, long long d = 1) {
    freq[d]++;
    for (long long x : v[z])
      if (x ^ p) dfs(x, z, d + 1);
  }
  void test_case() {
    cin >> n >> k;
    for (long long i = 1; i < n + 1; i++) v[i].clear();
    for (long long i = 1; i < n; i++) {
      long long x, y;
      cin >> x >> y;
      v[x].push_back(y), v[y].push_back(x);
    }
    if (k == 2) {
      cout << n * (n - 1) / 2 << "\n";
      return;
    }
    long long ans = 0;
    for (long long center = 1; center < n + 1; center++) {
      vector<vector<long long>> cnt(n + 1);
      for (long long x : v[center]) {
        memset(freq, 0, sizeof freq);
        dfs(x, center);
        for (long long j = 0; j < n; j++) {
          if (freq[j]) cnt[j].push_back(freq[j]);
        }
      }
      for (long long dep = 0; dep < n; dep++) {
        vector<long long> dp(k + 1);
        dp[0] = 1;
        for (long long x : cnt[dep]) {
          for (long long j = k - 1; j >= 0; j--) {
            dp[j + 1] = (dp[j + 1] + dp[j] * x) % 1000000007;
          }
        }
        ans = (ans + dp[k]) % 1000000007;
      }
    }
    cout << ans << "\n";
  }
  signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long t = 1;
    cin >> t;
    for (long long i = 0; i < t; i++) {
      test_case();
    }
    return 0;
  }
  