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;
  vector<string> vec_splitter(string s) {
    for (char& c : s) c = c == ',' ? ' ' : c;
    stringstream ss;
    ss << s;
    vector<string> res;
    for (string z; ss >> z; res.push_back(z))
      ;
    return res;
  }
  void debug_out(vector<string> args, int idx) { cerr << endl; }
  template <typename Head, typename... Tail>
  void debug_out(vector<string> args, int idx, Head H, Tail... T) {
    if (idx > 0) cerr << ", ";
    stringstream ss;
    ss << H;
    cerr << args[idx] << " = " << ss.str();
    debug_out(args, idx + 1, T...);
  }
  void localTest() {}
  const long long MOD = 1e9 + 7, N = 105;
  vector<long long> V[N];
  long long dp[N][N], level[N];
  void dfs(long long node, long long p, long long l) {
    level[l]++;
    for (auto child : V[node]) {
      if (child == p) {
        continue;
      }
      dfs(child, node, l + 1);
    }
    return;
  }
  void add(long long& x, long long y) {
    x += y;
    x %= MOD;
    return;
  }
  long long mul(long long x, long long y) {
    x %= MOD;
    y %= MOD;
    return (x * y) % MOD;
  }
  void solver() {
    int n, k;
    cin >> n >> k;
    for (int i = 1; i <= n; ++i) {
      V[i].clear();
    }
    for (int i = 1; i < n; ++i) {
      int 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 (int root = 1; root <= n; ++root) {
      if ((int)V[root].size() < k) {
        continue;
      }
      memset(dp, 0, sizeof dp);
      for (int i = 1; i <= n; ++i) {
        dp[i][0] = 1;
      }
      for (auto u : V[root]) {
        memset(level, 0, sizeof level);
        dfs(u, root, 1);
        for (int i = 1; i <= n; ++i) {
          for (int j = k - 1; j >= 0; --j) {
            add(dp[i][j + 1], mul(dp[i][j], level[i]));
          }
        }
      }
      for (int i = 1; i <= n; ++i) {
        add(ans, dp[i][k]);
      }
    }
    cout << ans << "\n";
    return;
  }
  int main() {
    localTest();
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin >> t;
    while (t--) {
      solver();
    }
    return 0;
  }
  