description: Moamen was drawing a grid of n rows and 10^9 columns containing only digits 0 and 1. Ezzat noticed what Moamen was drawing and became interested in the minimum number of rows one needs to remove to make the grid beautiful.
  
  A grid is beautiful if and only if for every two consecutive rows there is at least one column containing 1 in these two rows.
  
  Ezzat will give you the number of rows n, and m segments of the grid that contain digits 1. Every segment is represented with three integers i, l, and r, where i represents the row number, and l and r represent the first and the last column of the segment in that row.
  
  For example, if n = 3, m = 6, and the segments are (1,1,1), (1,7,8), (2,7,7), (2,15,15), (3,1,1), (3,15,15), then the grid is:
  
  <image>
  
  Your task is to tell Ezzat the minimum number of rows that should be removed to make the grid beautiful.
  
  Input
  
  The first line contains two integers n and m (1 ≤ n, m ≤ 3⋅10^5).
  
  Each of the next m lines contains three integers i, l, and r (1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ 10^9). Each of these m lines means that row number i contains digits 1 in columns from l to r, inclusive.
  
  Note that the segments may overlap.
  
  Output
  
  In the first line, print a single integer k — the minimum number of rows that should be removed.
  
  In the second line print k distinct integers r_1, r_2, …, r_k, representing the rows that should be removed (1 ≤ r_i ≤ n), in any order.
  
  If there are multiple answers, print any.
  
  Examples
  
  Input
  
  
  3 6
  1 1 1
  1 7 8
  2 7 7
  2 15 15
  3 1 1
  3 15 15
  
  
  Output
  
  
  0
  
  
  Input
  
  
  5 4
  1 2 3
  2 4 6
  3 3 5
  5 1 1
  
  
  Output
  
  
  3
  2 4 5
  
  Note
  
  In the first test case, the grid is the one explained in the problem statement. The grid has the following properties: 
  
    1. The 1-st row and the 2-nd row have a common 1 in the column 7. 
    2. The 2-nd row and the 3-rd row have a common 1 in the column 15. 
  
  As a result, this grid is beautiful and we do not need to remove any row.
  
  In the second test case, the given grid is as follows: 
  
  <image>
solution: #include <bits/stdc++.h>
  using namespace std;
  using uint = unsigned int;
  using ll = long long;
  using ull = unsigned long long;
  using pii = pair<int, int>;
  using pli = pair<ll, int>;
  using pil = pair<int, ll>;
  using pll = pair<ll, ll>;
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    auto solve = [&]() {
      int N, M;
      cin >> N >> M;
      vector<tuple<int, int, int>> events;
      for (int i = 0; i < M; i++) {
        int p, l, r;
        cin >> p >> l >> r;
        p--, r++;
        events.emplace_back(l, 1, p);
        events.emplace_back(r, 0, p);
      }
      sort(events.begin(), events.end());
      map<int, int> cnt;
      vector<pii> edges;
      for (auto [x, op, i] : events) {
        if (op == 1) {
          auto it = cnt.lower_bound(i);
          if (it == cnt.end() || it->first != i) {
            if (it != cnt.begin()) {
              edges.emplace_back(prev(it)->first, i);
            }
            if (it != cnt.end()) {
              edges.emplace_back(i, it->first);
            }
          }
          cnt[i]++;
        } else {
          auto it = cnt.find(i);
          it->second--;
          if (it != cnt.begin() && next(it) != cnt.end()) {
            edges.emplace_back(prev(it)->first, next(it)->first);
          }
          if (it->second == 0) {
            cnt.erase(it);
          }
        }
      }
      sort(edges.begin(), edges.end());
      vector<int> dp(N, 1), p(N, -1);
      for (auto [u, v] : edges) {
        if (dp[v] < dp[u] + 1) {
          dp[v] = dp[u] + 1;
          p[v] = u;
        }
      }
      vector<int> keep(N);
      int pnt = max_element(dp.begin(), dp.end()) - dp.begin();
      while (pnt >= 0) {
        keep[pnt] = 1;
        pnt = p[pnt];
      }
      vector<int> seq;
      for (int i = 0; i < N; i++) {
        if (!keep[i]) {
          seq.push_back(i);
        }
      }
      cout << int(seq.size()) << '\n';
      ;
      for (int i = 0; i < int(seq.size()); i++) {
        cout << seq[i] + 1 << " \n"[i == int(seq.size()) - 1];
      }
    };
    int t = 1;
    while (t--) {
      solve();
    }
    return 0;
  }
  