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;
  void main0();
  int main() {
    clock_t start, end;
    ios::sync_with_stdio(false);
    cin.tie(0);
    main0();
    return 0;
  }
  const int dx[8] = {0, 1, -1, 0, 1, 1, -1, -1};
  const int dy[8] = {1, 0, 0, -1, 1, -1, -1, 1};
  const int N = 105;
  const int M = 1e5;
  const int INF = 0x3f3f3f3f;
  const long long INFF = 0x3f3f3f3f3f3f3f3f;
  const int mod = 1e9 + 7;
  const double eps = 1e-6;
  const double Pi = acos(-1.0);
  mt19937 rnd(
      (unsigned int)chrono::steady_clock::now().time_since_epoch().count());
  long long n, k;
  long long ans;
  vector<int> e[N];
  int deg[N];
  int dep[N][N];
  int cnt[N][N];
  void dfs(int rt, int u, int fa) {
    dep[rt][u] = dep[rt][fa] + 1;
    cnt[rt][dep[rt][u]]++;
    for (int v : e[u])
      if (v != fa) {
        dfs(rt, v, u);
      }
  }
  long long dp[N][N];
  void check(int u) {
    for (int v : e[u]) {
      dep[v][u] = 0;
      for (int i = 1; i <= n; ++i) {
        cnt[v][i] = 0;
      }
      dfs(v, v, u);
    }
    for (int d = 1; d <= n; ++d) {
      memset(dp, 0, sizeof dp);
      dp[0][0] = 1;
      for (int i = 1; i <= e[u].size(); ++i) {
        for (int j = 0; j <= k; ++j) {
          dp[i][j] += dp[i - 1][j];
        }
        for (int j = 1; j <= k; ++j) {
          dp[i][j] += dp[i - 1][j - 1] * cnt[e[u][i - 1]][d] % mod;
          dp[i][j] %= mod;
        }
      }
      ans += dp[e[u].size()][k];
      ans %= mod;
    }
  }
  void main0() {
    int T;
    cin >> T;
    while (T--) {
      ans = 0;
      cin >> n >> k;
      for (int i = 1; i <= n; ++i) {
        e[i].clear();
        deg[i] = 0;
      }
      for (int i = 1; i <= n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        deg[u]++, deg[v]++;
        e[u].push_back(v);
        e[v].push_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
        continue;
      }
      for (int i = 1; i <= n; ++i) {
        if (deg[i] >= k) check(i);
      }
      cout << ans << endl;
    }
  }
  