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>
  #pragma GCC optimize("Ofast")
  #pragma GCC optimization("unroll-loops")
  long long md = 1000000007;
  using namespace std;
  template <typename T>
  T pw(T a, T b) {
    T c = 1, m = a;
    while (b) {
      if (b & 1) c = (c * m);
      m = (m * m), b /= 2;
    }
    return c;
  }
  template <typename T>
  T ceel(T a, T b) {
    if (a % b == 0)
      return a / b;
    else
      return a / b + 1;
  }
  template <typename T>
  T my_log(T n, T b) {
    T i = 1, ans = 0;
    while (1) {
      if (i > n) {
        ans--;
        break;
      }
      if (i == n) break;
      i *= b, ans++;
    }
    return ans;
  }
  template <typename T>
  T gcd(T a, T b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long pwmd(long long a, long long b) {
    long long c = 1, m = a;
    while (b) {
      if (b & 1) c = (c * m) % md;
      m = (m * m) % md;
      b /= 2;
    }
    return c;
  }
  long long modinv(long long n) { return pwmd(n, md - 2); }
  long long inverse(long long i) {
    if (i == 1) return 1;
    return (md - ((md / i) * inverse(md % i)) % md + md) % md;
  }
  bool sortbysec(const pair<long long, long long> &a,
                 const pair<long long, long long> &b) {
    return (a.second < b.second);
  }
  long long t, n, k, x, y, h = 0;
  vector<long long> adj[101], vis(101), dist(101);
  long long dis[101][101];
  void dfs(long long u) {
    vis[u] = 1;
    for (auto s : adj[u]) {
      if (vis[s]) continue;
      dist[s] = dist[u] + 1;
      dis[dist[s]][h]++;
      dfs(s);
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    ;
    cin >> t;
    while (t--) {
      cin >> n >> k;
      for (long long i = 0; i < n + 1; i++) adj[i].clear();
      for (long long i = 0; i < n - 1; i++)
        cin >> x >> y, adj[x].push_back(y), adj[y].push_back(x);
      if (k == 2) {
        cout << (n * (n - 1)) / 2 << "\n";
        continue;
      }
      long long ans = 0;
      for (long long i = 1; i < n + 1; i++) {
        for (long long j = 1; j < n + 1; j++) vis[j] = 0;
        if (adj[i].size() < k) continue;
        long long o = adj[i].size();
        h = 0;
        for (long long j = 0; j < n + 1; j++)
          for (long long p = 0; p < o; p++) dis[j][p] = 0;
        for (auto w : adj[i]) {
          dist[w] = 1;
          dis[1][h]++;
          for (long long p = 0; p < n + 1; p++) vis[p] = 0;
          vis[i] = 1;
          dfs(w), h++;
        }
        for (long long j = 1; j < n + 1; j++) {
          vector<long long> v;
          for (long long p = 0; p < o; p++) {
            if (dis[j][p]) v.push_back(dis[j][p]);
          }
          if (v.size() < k) continue;
          long long g = v.size();
          long long dp[g][k + 1];
          for (long long dd = 0; dd < g; dd++)
            for (long long tt = 0; tt < k + 1; tt++) dp[dd][tt] = 0;
          dp[0][0] = 1, dp[0][1] = v[0];
          for (long long dd = 1; dd < g; dd++) {
            dp[dd][0] = 1;
            for (long long tt = 1; tt < k + 1; tt++) {
              dp[dd][tt] += dp[dd - 1][tt];
              dp[dd][tt] %= md;
              dp[dd][tt] += (dp[dd - 1][tt - 1] * v[dd]) % md;
              dp[dd][tt] %= md;
            }
          }
          ans = (ans + dp[g - 1][k]) % md;
        }
      }
      cout << ans << "\n";
    }
  }
  