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;
  const int M = 1e9 + 7;
  long long sumproduct(vector<vector<int> > &cd, int k) {
    long long res = 0;
    for (auto a : cd) {
      int m = a.size();
      if (m < k) continue;
      vector<vector<long long> > dp(m, vector<long long>(k + 1));
      dp[0][1] = a[0];
      for (int i = 0; i < m; ++i) {
        dp[i][0] = 1;
      }
      for (int i = 1; i < m; ++i) {
        for (int j = 1; j <= k; ++j) {
          dp[i][j] = dp[i - 1][j] + 1LL * a[i] * dp[i - 1][j - 1];
          dp[i][j] %= M;
        }
      }
      res += dp[m - 1][k];
      res %= M;
    }
    return res;
  }
  void bfs(vector<vector<int> > &e, vector<int> &d, int u) {
    queue<int> q;
    q.push(u);
    d[u] = 1;
    while (!q.empty()) {
      int u = q.front();
      q.pop();
      for (int v : e[u]) {
        if (d[v] > 0) continue;
        d[v] = d[u] + 1;
        q.push(v);
      }
    }
  }
  int main() {
    int t;
    cin >> t;
    while (t--) {
      int n, k;
      cin >> n >> k;
      vector<vector<int> > e(n + 1);
      for (int i = 0; i < n - 1; ++i) {
        int u, v;
        cin >> u >> v;
        e[u].push_back(v);
        e[v].push_back(u);
      }
      if (k == 2) {
        cout << n * (n - 1) / 2 << endl;
        continue;
      }
      long long res = 0;
      for (int u = 1; u <= n; ++u) {
        vector<vector<int> > cd(n + 1);
        for (int v : e[u]) {
          vector<int> d(n + 1);
          d[u] = n + 1;
          bfs(e, d, v);
          d[u] = 0;
          vector<int> cnt(n + 1);
          for (int i = 1; i <= n; ++i) {
            ++cnt[d[i]];
          }
          for (int i = 1; i <= n; ++i) {
            if (cnt[i] > 0) cd[i].push_back(cnt[i]);
          }
        }
        res += sumproduct(cd, k);
        res %= M;
      }
      cout << res << endl;
    }
    return 0;
  }
  