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;
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, m;
    cin >> n >> m;
    vector<pair<int, int>> endpoint;
    for (int i = 0; i < m; i++) {
      int id, l, r;
      cin >> id >> l >> r;
      endpoint.emplace_back(l, id);
      endpoint.emplace_back(r + 1, (~id));
    }
    sort(endpoint.begin(), endpoint.end());
    vector<vector<int>> G(n + 1);
    map<int, int> mp;
    int beg = 0;
    while (beg < (int)endpoint.size()) {
      int end = beg;
      while (end + 1 < (int)endpoint.size() &&
             endpoint[end + 1].first == endpoint[end].first) {
        ++end;
      }
      for (int i = beg; i <= end; i++) {
        int id = endpoint[i].second;
        if (id >= 0) {
          auto it = mp.lower_bound(id);
          if (it != mp.end() && it->first == id) {
            it->second += 1;
            continue;
          }
          if (it != mp.begin()) {
            int x = prev(it)->first;
            G[x].push_back(id);
          }
          if (it != mp.end()) {
            int y = it->first;
            G[id].push_back(y);
          }
          mp[id] = 1;
        } else {
          id = ~id;
          auto it = mp.find(id);
          assert(it != mp.end());
          if (it->second > 1) {
            it->second -= 1;
            continue;
          }
          if (it != mp.begin() && next(it) != mp.end()) {
            int x = prev(it)->first;
            int y = next(it)->first;
            G[x].push_back(y);
          }
          mp.erase(it);
        }
      }
      beg = end + 1;
    }
    vector<int> dp(n + 1, 1);
    vector<int> pa(n + 1, -1);
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < (int)G[i].size(); j++) {
        if (dp[i] + 1 > dp[G[i][j]]) {
          dp[G[i][j]] = dp[i] + 1;
          pa[G[i][j]] = i;
        }
      }
    }
    int pos = (int)(max_element(dp.begin() + 1, dp.end()) - dp.begin());
    int ans = n;
    vector<bool> remain(n + 1, 0);
    while (pos > 0) {
      remain[pos] = 1;
      ans--;
      pos = pa[pos];
    }
    cout << ans << '\n';
    for (int i = 1; i <= n; i++) {
      if (!remain[i]) {
        cout << i << " ";
      }
    }
    cout << '\n';
    return 0;
  }
  