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;
  const int MX = 3e5 + 10;
  const int INF = 1e9;
  bool vis[MX];
  set<int> s;
  int cnt[MX], N, M, dp[MX], pre[MX];
  vector<pair<int, int> > rg;
  vector<int> adj[MX];
  void chmax(int& a, int b) {
    if (b > a) a = b;
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> N >> M;
    for (int i = 0; i < M; i++) {
      int x, l, r;
      cin >> x >> l >> r;
      rg.push_back({l, x});
      rg.push_back({r + 1, -x});
    }
    sort(rg.begin(), rg.end());
    set<int>::iterator it;
    s.insert(0);
    s.insert(N + 1);
    for (auto p : rg) {
      int id = p.second > 0 ? p.second : -p.second;
      if (p.second > 0) {
        cnt[id]++;
        if (cnt[id] == 1) {
          s.insert(id);
          it = s.find(id);
          it--;
          adj[id].push_back((*it));
          it++;
          it++;
          adj[(*it)].push_back(id);
        }
      } else {
        cnt[id]--;
        if (cnt[id] == 0) s.erase(id);
      }
    }
    for (int i = 1; i <= N + 1; i++) dp[i] = -INF;
    for (int i = 1; i <= N + 1; i++) {
      for (auto j : adj[i]) {
        if (dp[j] + 1 > dp[i]) {
          dp[i] = dp[j] + 1;
          pre[i] = j;
        }
      }
    }
    cout << N + 1 - dp[N + 1] << '\n';
    int curr = N + 1;
    while (curr != 0) {
      vis[curr] = 1;
      curr = pre[curr];
    }
    for (int i = 1; i <= N; i++)
      if (!vis[i]) cout << i << " ";
    cout << '\n';
  }
  