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 <class A, class B>
  ostream& operator<<(ostream& out, const pair<A, B>& p) {
    return out << "(" << p.first << ", " << p.second << ")";
  }
  template <class A>
  ostream& operator<<(ostream& out, const vector<A>& v) {
    out << "[";
    for (int i = 0; i < v.size(); ++i) {
      if (i) out << ", ";
      out << v[i];
    }
    return out << "]";
  }
  const int N = 106;
  int n, k;
  const long long MOD = 1e9 + 7;
  long long mul(long long o1, long long o2) { return o1 * o2 % MOD; }
  vector<int> adj[N];
  vector<int> dis[N];
  void init() {
    for (int i = (0); i < (n); ++i) {
      vector<int>().swap(adj[i]);
      vector<int>().swap(dis[i]);
      dis[i].resize(N);
    }
  }
  vector<int> cnt;
  long long getDP(int sub) {
    long long dp[sub + 5][k + 5];
    for (int i = (0); i < (sub + 2); ++i) {
      for (int j = (0); j < (k + 1); ++j) {
        dp[i][j] = 0;
      }
    }
    dp[0][0] = 1;
    for (int i = (0); i < (sub); ++i) {
      for (int j = 0; j <= k; ++j) {
        dp[i + 1][j] += dp[i][j];
        dp[i + 1][j] %= MOD;
        dp[i + 1][j + 1] += mul(dp[i][j], cnt[i]);
        dp[i + 1][j + 1] %= MOD;
      }
    }
    return dp[sub][k];
  }
  void sol() {
    cin >> n >> k;
    for (int i = (0); i < (n); ++i) {
      dis[i].resize(n);
    }
    for (int i = (1); i < (n); ++i) {
      int x, y;
      cin >> x >> y;
      --x;
      --y;
      adj[x].push_back(y);
      adj[y].push_back(x);
    }
    long long ans = 0;
    if (k == 2) {
      ans = (1ll * n * (n - 1) / 2) % MOD;
    } else {
      bool vis[n];
      for (int center = (0); center < (n); ++center) {
        for (int i = (0); i < (n); ++i) {
          vis[i] = false;
        }
        vis[center] = true;
        vector<pair<int, int> > layer;
        vector<int>(n).swap(cnt);
        int sz = adj[center].size();
        for (int i = (0); i < (sz); ++i) {
          vis[adj[center][i]] = true;
          layer.emplace_back(i, adj[center][i]);
          cnt[i] = 1;
        }
        while (!layer.empty()) {
          ans += getDP(sz);
          ans %= MOD;
          vector<pair<int, int> > next_layer;
          for (pair<int, int> e : layer) {
            --cnt[e.first];
            for (int to : adj[e.second]) {
              if (vis[to]) continue;
              ++cnt[e.first];
              next_layer.emplace_back(e.first, to);
              vis[to] = true;
            }
          }
          swap(next_layer, layer);
          vector<pair<int, int> >().swap(next_layer);
        }
      }
    }
    cout << ans << endl;
    init();
  }
  int main(int argc, char const* argv[]) {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int t;
    cin >> t;
    for (; t; --t) {
      sol();
    }
    return 0;
  }
  